经过了一段时间的努力，我再Pku上也算是有了一个阶段性的总结拉，下面是我就这段时间搞ACM来的一些代码的总结，具体的一些题目类型的总结看本Blog的相关文章。

huicpc26 ACM_PKU 代码总结

1、DP（动态规划）
/*1080-HumanGeneFunctions.cpp*/
观察题目给出的一个最优解：
AGTGATG
-GTTA-G
将其从某一处切开，如果左边部分的分值不是最大，那么将其进行调整，使其分值变大，
则整个解分值变大，与已知的最优矛盾。所以左边部分的分值必是最大。
同理，右边也是。可见满足最优子结构的性质。考虑使用DP：
设两个DNA序列分别为s1，s2，长度分别为len1，len2，score为分值表。
f[i,j]表示子串s1[1..i]和s2[1..j]的分值。考虑一个f[i,j]，我们有：
  1.s1取第i个字母，s2取“-”：f[i-1,j] + score[s1[i],''-'']
  2.s1取“-”，s2取第j个字母：f[i,j-1] + score[''-'',s2[j]]
  3.s1取第i个字母，s2取第j个字母：f[i-1,j-1] + score[s1[i],s2[j]]
  即f[i,j] = max(f[i-1,j] + score[s1[i],''-''], f[i,j-1] + score[''-'',s2[j]], f[i-1,j-1] + score[s1[i],s2[j]]);
然后考虑边界条件，这道题为i或j为0的情况。
当i=j=0时，即为f[0,0]，这是在计算f[1,1]时用到的，根据f[1,1] = f[0,0] + score[s1[i], s2[j]]，明显有f[0,0] = 0。
当i=0时，即为f[0,1..len2]，有了f[0,0]，可以用f[0,j] = f[0,j-1] + table[''-'',s2[j]]来计算。
当j=0时，即为f[1..len1,0]，有了f[0,0]，可以用f[i,0] = f[i-1,0] + table[s1[i],''-'']来计算。
至于计算顺序，只要保证计算f[i,j]的时候，使用到的f[i-1,j]，f[i,j-1]，f[i-1,j-1]都计算出来了就行了。
所谓划分阶段也就是为了达到这个目的。这样我们使用一个二重循环就可以了。*/
#include
#include
#include
using namespace std;
#define MAX(a,b,c) (a>b?a:b)>c?(a>b?a:b):c
int ctoi(char a){
    int b;
    if(a==''A'')        b = 0;
    if(a==''C'')        b = 1;
    if(a==''G'')        b = 2;
    if(a==''T'')        b = 3;
    if(a==''-'')        b = 4;
    return b;
}
int main()
{
    int t,j,k,m,n;
    int f1,f2,f3;
    int f[101][101];
int arr[5][5]={{5,-1,-2,-1,-3},{-1,5,-3,-2,-4},{-2,-3,5,-2,-2},{-1,-2,-2,5,-1},{-3,-4,-2,-1,0}};
    string a,b;
    cin>>t;
    while(t--){
        j = k = 0;
        memset(f,0,sizeof(f));
        cin>>m>>a;
        cin>>n>>b;
        for(j=0;j<=m;j++)
        {
            for(k=0;k<=n;k++)
            {
                if(j == 0 && k == 0)
                {
                    f[j][k] = 0;
                }
                else if(j==0)
                {
                    f[j][k] = f[j][k-1] + arr[ctoi(''-'')][ctoi(b[k-1])];
                }
                else if(k==0)
                {
                    f[j][k] = f[j-1][k] + arr[ctoi(a[j-1])][ctoi(''-'')];
                }
                else
                {
                    f1 = f[j-1][k]   + arr[ctoi(a[j-1])][ctoi( ''-'')];
                    f2 = f[j][k-1]   + arr[ctoi( ''-'')][ctoi(b[k-1])];
                    f3 = f[j-1][k-1] + arr[ctoi(a[j-1])][ctoi(b[k-1])];

                    f[j][k] = MAX(f1,f2,f3);                
                }
            }
        }
        cout<     }
    return 0;
}
2、/*1088-滑雪.cpp*/
#include
using namespace std;
const int highest = 100000;
int high[100][100];
bool travel[100][100];
int value[100][100];
int r, c;
const int dir[4][2] = {0, 1, 1, 0, -1, 0, 0, -1};
int find_rout(int x, int y){
    int i, temp;
    if(x < 0 || y < 0 || x >= r || y >= c)
        return 0;
    if(travel[x][y])
        return value[x][y];
    int max_rout = 1;
    for(i = 0; i < 4; ++i){
        if(high[x][y] > high[x + dir[i][0]][y + dir[i][1]])
        {
            temp = find_rout(x + dir[i][0], y + dir[i][1]);
            if(max_rout < temp + 1)
                max_rout = temp + 1;
        }
    }
    travel[x][y] = true;
    value[x][y] = max_rout;
    return max_rout;
}
int main(){
    cin >> r >> c;
    int i, j;
    int max_rout = 1;
    for(i = 0; i < r; ++i)
        for(j = 0; j < c; ++j){
            travel[i][j] = false;
            cin >> high[i][j];
        }
    for(i = 0; i < r; ++i)
        for(j = 0; j < c; ++j)    {
            int temp = find_rout(i, j);
            if(temp > max_rout)
                max_rout = temp;
        }
    cout << max_rout << endl;
    return 0;
}
3、精度运算
/*2262-Goldbach''sConjecture.cpp*/
#include
#include
void doRun(){
    int i,j,sum,a;
    long t;
    while(cin>>a){
        t = 0;
        if(a== 0||a>10000)
            break;
        sum = 0;
        i = 1;
        while(1){
            sum += i;
            if(sum>=a){
                t = a-(sum-i);
                break;
            }
            i++;
        }
        t = t*i;
        for(j=1;j             t += j*j;
        }
        cout<         printf("%ld %ld",a,b);
        if( a>b ){
            n = a ;
            a = b ;
            b = n ;
        }
        s = 0;
        for(i=a; i<=b; i++){
            count = 1;
            temp = i;
            while(temp !=1){
                if(temp%2 == 1)temp = 3*temp + 1;
                else          temp = temp/2;
                count++;
            }
            if(count>s)        s = count;
        }
        printf(" %ld\n",s);
    }
}
int main(){
    doRun();
    return 0;
}
5、/*2080-Calendar.cpp*/
#include
#include
using namespace std;
int mday[2][13] = {  0,31,28,31,30,31,30,31,31,30,31,30,31,0,31,29,31,30,31,30,31,31,30,31,30,31 };
char week[7][20] = { "Saturday","Sunday","Monday","Tuesday","Wednesday","Thursday","Friday" };
int main ()
{
    int n;
    int s, y, m, d, t;
    while ( 1 ) {
        cin >> n ;
        if ( n < 0 )
            break;
        d = n % 7;
        n ++ ;
        for ( y = s = 0 ; s < n; s += t, y ++ )
            if ( y % 400 == 0 || y % 4 == 0 && y % 100 )
                t = 366;
            else
                t = 365;
            y -- ; s -= t; n -= s;
            cout << y+2000 << ''-'';
            t -= 365;
            for ( m = 0, s = 0; s < n; m ++,s += mday[t][m]  );
            cout << setw(2) << setfill(''0'') << m << ''-'';
            s -= mday[t][m]; n -= s;
            cout << setw(2) << setfill(''0'') << n << '' '';
            cout << week[d] << endl;
    }
    return 0;
}
6、/*1298-TheHardestProblemEver.cpp*/
#include
#include
using namespace std;
void main()
{
    string S1("ABCDEFGHIJKLMNOPQRSTUVWXYZ");
    string S2("VWXYZABCDEFGHIJKLMNOPQRSTU");
    string start,end,over("ENDOFINPUT");
    string str;
    char ch;
    int i,j;
    while(1)
    {
        getline(cin, start);
        if(start.compare(over)==0)break;/*        
        while((ch = getchar())!=''\n'')
        {
            if(ch<=''Z''&&ch>=''A'')
            {
                j=S1.find(ch);
                ch = S2[j];
            }
            cout<         }
        cout<         getline(cin, str);
        for(i=0;i         {
            if(str[i]<=''Z''&&str[i]>=''A'')
            {
                j=S1.find(str[i]);
                str[i]=S2[j];
            }
        }
        cout<         getline(cin, end);
    }
}
7、/*1745-Divisibility.cpp*/
#include
#include
int d[105], e[105], a, n, k;
int main ()
{
    int i, p, q;
    scanf ( "%d%d", &n, &k );
    memset ( d, 0, k*sizeof(int) );
    scanf ( "%d", &a );
    a = a%k;
    if ( a < 0 )    a += k;
    d[a] = 1;
    while ( --n )
    {
        scanf ( "%d", &a );
        p = a%k;
        q = (-a)%k;
        if ( p < 0 ) p += k;
        if ( q < 0 ) q += k;
        memset ( e, 0, k*sizeof(int) );
        for ( i = 0; i < k; i ++, p ++, q ++ )
        {
            if ( d[i] == 1 )
            {
                e[p%k] = e[q%k] = 1;
            }
        }
        memcpy ( d, e, k*sizeof(int) );
    }
    printf ( d[0] ? "Divisible\n" : "Not divisible\n" );
    return 0;
}
8、/*2249-BinomialShowdown.cpp*/
#include
#include
using namespace std;
int main()
{
   int n,k;
   double temp;
   while(cin>>n>>k)
   {
      if(n==0&&k==0) break;
      else
      {
         temp=1;
         if(n-k>k)
         {
            for(int i=n-k+1;i<=n;++i)
               temp*=i;
            for(int j=1;j<=k;++j)
               temp/=j;
            cout<          }
         else
         {
            for(int i=k+1;i<=n;++i)
               temp*=i;
            for(int j=1;j<=n-k;++j)
               temp/=j;
            cout<          }
      }
   }
}
9、/*2309-BST.c*/
#include
void doRun()
{
    int n;
    long long i,j,k,s;
    long long a;
    scanf("%d",&n);
    for(i=0;i     {
        scanf("%I64d",&a);
        s = a;
        if( (s&1)==1)
            printf("%I64d %I64d\n",s,s);
        else
        {
            j=0;
            while(1)
            {
                if((s&1) == 1)
                    break;
                s >>= 1;
                j++;
            }
            k = 1;
            k <<= j;
            printf("%I64d %I64d\n",a-k+1,a+k-1);
        }
    }
}
int main()
{
    doRun();
    return 0;
}
10、/*1767-WhichisNext.cpp*/
#include
#include
#include
using namespace std;

long NextTreeIdentify(string t)
{
    string s,t2,t001("001");
    s = t;
    long res;
    int node,i,posof0,posof001;
    bool end = true;
    for(i=0;i<=s.size()-3;i++)
    {
        t2 = s.substr(i,3);
        if(t2.compare(t001)==0)
        {
            posof001 = i;
            break;
        }
    }
    end = true;
    for(i=posof001+3;i     {
        if(s[i] != ''1'')
        {
            end = false;
            break;
        }
    }
    if(!end)
    {
        /*s.erase(posof001+1,2);
        posof0 = s.find_first_of("0",posof001+1);
        s.insert(posof0+1,"1");
        s.insert(posof0+1,"0");*/
        s.replace(posof001,3,"0");
        posof0 = s.find_first_of("0",posof001+1);
        s.replace(posof0,1,"001");
    }
    else
    {
        s[0] = ''0'';
        for(i=1;i         {
            if(i%2 == 0)
                s[i] = ''1'';
            else
                s[i] = ''0'';
        }
    }
    i = s.size()-1;
    res = 0;
    while(i>=2)
    {
        node = s[i]-''0'';
        if(node == 1)
            res +=  node<         i--;
    }
    return res;
}
void doRun()
{
    long res,n,t;
    int node;
    stack tree;
    string s;
    scanf("%ld",&n);
    if(n == 0||n == 4)
    {
        res = n;
    }
    else
    {
        t = n;
        while(1)
        {
            node = t % 2;
            tree.push(node);
            if (t<=1)break;
            t>>=1;
        }
        if(tree.size()>30)
            return;
        while(!tree.empty())
        {
            s.insert(0,(char*)(tree.top()+''0''));
            tree.pop();
        }
        res = NextTreeIdentify(s);
    }
    printf("%ld\n",res);
}
int main()
{
    doRun();
    return 0;
}
11、/*2229-Sumsets.cpp*/
#include
#define M 1000001
int b[M];
void doRun()
{
    int i;
    b[1] = 1;
    b[2] = 2;
    n = 0;
    for(i=2;i     {
        b[i] = b[i++-1];
        b[i] = (b[i-2]+b[i/2])%1000000000;
    }
    while(scanf("%d",&i)!=EOF)
            printf("%d\n",b[i]);
}
int main()
{
    doRun();
    return 0;
}
12、/*2533-LongestOrderedSubsequence.c*/
#include
void process(int *a, int sum)
{
   int i,j;
   int f[1000],res,max;
   for(i=1; i<=sum; i++)
        f[i]=0;
   res = 0;
   for(i=sum;i>0;i--)
   {
      max = 0;
      for(j=i+1; j<=sum; j++)
      {
          if(a[j]>a[i])
          {
              if(max               {
                  max = f[j];
              }    
          }
      }
      f[i] = max + 1;
      if(res       {
          res = f[i];
      }    
   }
   printf("%d",res);
}
main()
{
    int sum;
    int a[1000];
    int i;
    scanf("%d", &sum);
    for(i=1; i<=sum; i++)
        scanf("%d",a+i);
    process(a,sum);
    return 0;
}

13、/*2545-HammingProblem.c*/
#include
#include
#include
int isPrime(long long a)
{
    int flag=1 ;
    long long i;
    for(i=2;i     {
        if(a%i==0&&a!=2)
        {
            flag=0 ;
            break ;
        }
    }
    return flag ;
}
long long getMin3(long long a,long long b,long long c)
{
    long long t=a>b?b:a;
    return t>c?c:t ;
}
long long  getNum(long long  a,long long  b,long long  c,long long  pos)
{
    long long  p[3],i,j,count = 1;
    long long temp[3]= {0},min;
    long long * num = (long long *)malloc((pos+2)*sizeof(long long));
    p[0] = a;p[1] = b;p[2] = c;
    num[0] = 1;
    count = 1;
    while(count!=pos+1)
    {
        for(i=0;i<3;i++)
        {
            for(j=0;j             {
                temp[i] = num[j]*p[i];
                if (temp[i]>num[count-1])
                    break;
            }
        }
        min = getMin3(temp[0],temp[1],temp[2]);
        num[count]=min ;
        count++;
    }
    min = num[pos];
    return min;
}
int main()
{
    long long p1,p2,p3,i ;
    scanf("%I64d %I64d %I64d %I64d",&p1,&p2,&p3,&i);\
    if(isPrime(p1)==0||isPrime(p2)==0||isPrime(p3)==0||p1==p2||p2==p3||p1==p3)
        return 1;
    printf("%I64d",getNum(p1,p2,p3,i));
    getchar();
    return 0 ;
}
14、/*2567-CodetheTree.cpp*/
#include
#include
using namespace std;
struct node
{
    int parent;
    int child[51];
    int du;
}nd[51];
void doRun()
{
    int i,j,k,len,temp,root;
    char s[256],b[2];
    stack sp;
    int p[51];
    while(gets(s)!=NULL)
    {
        for(i=0;i<51;i++)
        {
            nd[i].du = 1;
            nd[i].parent = 0;
            for(j=0;j<51;j++)
                nd[i].child[j] = 0;
        }
        len = 0;
        //计算父亲节点和度
        sp.push((int)''('');
        for(i=1;i<(int)strlen(s);i++)
        {
            if(isdigit(s[i]))
            {
                len++;                
                j=0;
                while(s[i]!='' ''&& s[i]!='')'')
                {
                    b[j++] = s[i];
                    i++;
                }
                i--;
                b[j] = ''\0'';
                temp = atoi(b);
                sp.push(temp);
                if(i==1)
                {
                    root = temp;
                    nd[root].du--;
                }
            }
            else if(s[i] == ''('')
            {
                nd[sp.top()].du++;
                sp.push((int)''('');
            }
            else if(s[i] == '')'')
            {
                while(sp.top()!=(int)''('')
                {
                    temp = sp.top();
                    sp.pop();
                }
                sp.pop();
                if(!sp.empty())
                {
                    nd[temp].parent = sp.top();
                }
                else
                {
                    nd[temp].parent = 0;
                }
            }
            else
            {
                continue;
            }
        }
        if(len <= 1)
        {
            if(len==1)
                cout<             continue;
        }
        //求出孩子节点
        for(j=1;j<=len;j++)
        {
            for(i=1; i<=len; i++)
            {
                if(nd[i].parent!=0 && nd[i].parent == j)
                {
                    nd[j].child[i] = i;
                }
            }
        }
        //求出删除点的父亲节点
        k=0;
        for(i=1; i<=len; i++)
        {
            if(nd[i].du == 1)
            {
                if(nd[i].parent != 0)
                {
                    p[k++] = nd[i].parent;
                    nd[nd[i].parent].child[i] = 0;
                    nd[nd[i].parent].du--;
                }
                else
                {
                    temp = 0;
                    while(1)
                    {
                        if(nd[i].child[temp] != 0)
                        {
                            p[k++] = nd[i].child[temp];
                            nd[nd[i].child[temp]].parent = 0;
                            nd[nd[i].child[temp]].du--;
                            break;
                        }
                        temp++;
                    }
                }
                nd[i].du--;
                i = 0;
            }
        }
        //输出
        cout<         for(i=1;i         {
            cout<<" "<         }
        cout<     }
}
int main()
{
    doRun();
    return 0;
}
15、/*2568-DecodetheTree.c*/
#include
#include
char * toString(int a)
{
    char * b;
    b = (char*)malloc(2*sizeof(char));
    if(a>9)
    {
        b[0] = (int)(a/10)+''0'';
        b[1] = (a%10) + ''0'';
        b[2] = ''\0'';
    }
    else
    {
        b[0] = a + ''0'';
        b[1] = ''\0'';
    }
    return b;
}
int toInt(char *s)
{
    int a;
    if(strlen(s)>1)
    {
        a=(s[0]-''0'')*10+(s[1]-''0'');
    }
    else
        a = s[0]-''0'';
    return a;
}
int main()
{
    int num[51],f[51],i,j,k,count,a,flag;
    char s[256],temp[51][256],b[2];
    
    while(gets(s)!=NULL)
    {
        i=count=0;
        while(i<(int)strlen(s))
        {
            if(s[i] == '' '')
            {
                i++;
                continue;
            }
            else
            {
                j=0;
                while(s[i]!='' '')
                {
                    b[j++] = s[i];
                    i++;
                }
                b[j] = ''\0'';
                num[count++] = toInt(b);
            }
        }
        if(count == 0)
        {
            printf("(1)\n");
            continue;
        }        
        for(i=0;i<=count+1;i++)
        {
            strcpy(temp[i],toString(i));
        }
        for(i=0;i<=count+1;i++)
        {
            f[i] = 1;
        }
        for(i=0; i         {
            f[num[i]]++;
        }
        for(i=0;i         {
            a = num[i];
            j=1;
            while(j<=count)
            {
                if(f[j]==1)
                {   
                     f[j] = 0;
                     f[a]--;
                     break;
                }
                j++;
            }
            strcat(temp[a]," (");
            strcat(temp[a],temp[j]);
            strcat(temp[a],")");
        }
        printf("(%s)\n",temp[num[count-1]]);
    }
    return 0;
}
16、随机化问题
/*2576-TugofWar.c*/
#include
#include
#define MAX 100
int n , num[2] , sum[2] , in[2][MAX/2] ;
void input()
{
    int i ;
    for( i=num[0]=sum[0]=0 ; i     {
        scanf( "%d" , &in[0][ num[0] ] ) ;
        sum[0] += in[0][ num[0] ] ;
    }
    for( i=n/2,num[1]=sum[1]=0 ; i     {
        scanf( "%d" , &in[1][ num[1] ] ) ;
        sum[1] += in[1][ num[1] ] ;
    }
}
void process()
{
    int changed , i , j ;    
    input() ;    
    for( changed=0 ; ; changed=0 )
    {
        for( i=0 ; i             for( j=0 ; j                 {
                    sum[0] = sum[0]-in[0][i]+in[1][j] ;
                    sum[1] = sum[1]-in[1][j]+in[0][i] ;
                    in[0][i] ^= in[1][j] ^= in[0][i] ^= in[1][j] ;/* swap */
                    changed = 1 ;
                }

        if( !changed ) break ;
    }
    printf( "%d %d\n" , sum[0]sum[1]?sum[0]:sum[1] ) ;
}
int main()
{
    while( scanf( "%d" , &n )==1 )
        process() ;
    return 0 ;
}
17、/*2590-steps.cpp*/
#include
int main()
{
    long a,b,d;
    int n,i,c,step;
    cin>>n;
    for(i=0;i     {
        cin>>a>>b;
        d = b-a;
        c = 1;
        step = 0;
        while(1)
        {
            if (d < 2*c)
            {
                break;
            }
            else
            {
                d = d-2*c;
                step += 2;
                c++;
            }
        }
        if(d > c)
            step += 2;
        else if(d <= 0)
            step += 0;
        else
            step += 1;
        cout<     }
    return 0;
}
18、/*2605-Simplegameonagrid.cpp*/
#include
void doRun()
{
    int m,n;
    while(cin>>m>>n)
    {
        if(m         {
            m = m^n;
            n = n^m;
            m = m^n;
        }
        if(n==1)
        {
            cout<<(m+1)/2;
        }
        else if(n%3==0 || m%3==0)
        {
            cout<<2 ;
        }
        else
        {
            cout<< 1;
        }
        cout<     }
}
int main()
{
    doRun();
    return 0 ;
}
19、/*2615-Suffidromes.c*/
#include
char a[2000],b[2000];
int al,bl ;
int Works(char*s,int sl,int n)
{
    int i ;    
    for(i=n;i     {
        if(s[i]!=s[sl+n-i-1])
        {
            return 0 ;
        }
    }
    return 1 ;
}
main()
{
    int i,j,k,x,y,z,n ;
    char*s ;
    int sl ;    
    for(;;)
    {
        if(!gets(a)||!gets(b))break ;
        if(!strcmp(a,b))
        {
            printf("No Solution.\n");
            continue ;
        }
        al=strlen(a);
        bl=strlen(b);
        for(x=0;a[x]==b[x];x++);
        for(i=0;i<=al&&i<=bl;i++)
        {
            z=Works(a,al,i)+2*Works(b,bl,i);
            if(z==1||z==2||(z==3&&i>x))
            {
                switch(z)
                {
                    case 1 : s=a ; break ;
                    case 2 : s=b ; break ;
                    case 3 :
                    for(j=i-1;j>=0;j--)if(a[j]!=b[j])break ;
                    s=(a[j]                     break ;
                }
                for(j=i-1;j>=0;j--)putchar(s[j]);
                putchar(''\n'');
                break ;
            }
        }
        if(i>al||i>bl)
        {
            s=(i>al)?b:a ;
            sl=(i>al)?bl:al ;
            if(s[i-1]==''a''&&Works(s,sl,i))
            putchar(''b'');
            else
            putchar(''a'');
            s=(i>al)?a:b ;
            sl=(i>al)?al:bl ;
            for(j=sl-1;j>=0;j--)putchar(s[j]);
            putchar(''\n'');
        }
    }
}

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