博文

stu(1154)(2005-08-30 20:18:00)

摘要:#include<iostream.h>
int main()
{
    int k,i,j,t_case,t_piont,count;
    int x[1000],y[1000];
    cin>>t_case;
    if((t_case<1)&&(t_case>1000)) return 0;
    for(k=0;k<t_case;k++)
    {
        count=0;
        cin>>t_piont;
        for(i=0;i<t_piont;i++)
        {
            cin>>x[i]>>y[i];
      if((x[i]<-10000)&&(x[i]>10000)&&(y[i]<-10000)&&(y[i]>10000))
                return 0;
 &nbs......

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stu(1101)(2005-08-30 20:17:00)

摘要:#include<stdio.h>
#include<iostream.h>
int main()
{
  int n[15],i=0,j,twice=0;
  cin>>n[i];
  while(n[i]!=-1)
    {
      cin>>n[++i];
      if(n[i]==0)
    {
          cout<<twice<<endl;
      twice=0;
      i=0;
      cin>>n[i];
    }
     else
    {
      for(j=0;j<i;j++)
         {
        if(n[j]==2*n[i]||n[i]==2*n[j])
        twice++;
     &nbs......

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stu(1059)(2005-08-30 20:16:00)

摘要:#include"iostream.h"
void print(int *data,int n)
{
    int i;
    int a,b;
    
        
        //-----判断前面有没有相同的点,如果有,则从第二个起将所有的点前移,并将点的总数n减1-----
        while(data[2]==data[0]&&data[3]==data[1])
    {
        for(i=0;i<2*n-2;i++)    data[i]=data[i+2];     //前移        
        n--;                                   //总数减1
    }
    
 ......

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stu(1030)(2005-08-30 20:15:00)

摘要:这是汕头大学的第1030道题,答案:
#include<stdio.h>
#include<string.h>

int change(char C[],int k)
{
    int i;
    for(i=0;i<k;i++)
        {
            if(C[i]/10)
            {
                C[i]=C[i]%10;
                C[i+1]++;
            }
        }
    if(C[i])
        k++;
    return k;
}

int main()
{
    char A[52],B[52],C[52];
&nb......

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stu(1027)(2005-08-30 20:14:00)

摘要:#include<stdio.h>
#include<stdlib.h>

long check(long r)
{ long x=1;
  while((r/2)%2==0)
  { x++;
    r=r/2;
  }
  if(r!=1)
    return(0);
  else
    return(x);
}

long count(long n)
{ int k=1;
  while(!check(k*n+1))
   k++;
}

int main()
{ long n,x;
  while(scanf("%ld",&n)!=EOF)
  { if(n==1||n%2==0)
      printf("2^? mod %ld=1\n",n);
    else
    { x=count(n);
      printf("2^%ld mod %ld=1\n",x,n);
    }
  }
  return 0;
......

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stu(1021)Z(2005-08-30 20:12:00)

摘要:这题是汕头大学的1021道题:答案是:
#include<stdio.h>
#include<stdlib.h>

int change()
{ char x;
  int sum=0;
  scanf("%c",&x);
  if(x!='0')
  { do
  { sum=sum+x-48;
    scanf("%c",&x);
  }while(x>=48);
  }
  else
    return(0);
  return(sum);
}

int root(long x)
{ int sum=0;
  while(x>9)
  { sum=sum+x%10;
    x=x/10;
  }
  return(sum+x);
}

int main()
{ int x;
  x=change();
  while(x)
  {
    do
    { x=root(x);
    }while(x>9);
    printf("%d\n",x);
    x=change();
  }......

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pku(2591)(2005-08-28 14:10:00)

摘要:#include <iostream.h>
#include <stdio.h>
long a[10000002];
long min(long a,long b)
{
    if(a<b) return a;
    else
        return b;
}
int main()
{
    long i,j;
    long n,k;
        i=1,j=1;
        a[1]=1;
        for(k=2;k<=10000001;k++)
        {
            a[k]=min(2*a[i]+1,3*a[j]+1);
            if(2*a[i]+1==a[k])
                i++;
            if(3......

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jnu(String Matching)(2005-08-27 16:00:00)

摘要:#include <iostream.h>
#include <fstream.h>
ifstream f("String.in");
#define cin f
#include <string.h>
int main()
{
    char a[1000],b[1000],temp[1000];
    while(cin>>a)
    {
        int total=0;
        int sum,i,j,k;
        if(strcmp(a,"-1")==0) return 0;
        cin>>b;
        cout<<"appx("<<a<<","<<b<<") = ";
        if(strlen(a)<strlen(b))
        {
            strcpy(temp,a);
    &nb......

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jnu(Function Run Fun)(2005-08-27 16:00:00)

摘要:#include <math.h>
#include <iostream.h>
int w(int a,int b,int c)
{
    if(a<=0 || b<=0 || c<=0)
        return 1;
    else if(a>20 || b>20 || c>20)
        return w(20,20,20);
    else if(a==b && c==b)
        return (int)pow(2,a);
    else if((a<=b && a<=c) || (a<=c && b<=c) || (b>=a && b>=c))
        return (int)pow(2,a);
    else
    return w(a-1, b, c) + w(a-1, b-1, c) + w(a-1, b, c-1) - w(a-1, b-1, c-1) ;
}
int main()
{
    int a,b,c;
    while(cin>>a>>b>>c)
    {
 &n......

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jnu(Partial Sums)(2005-08-26 14:48:00)

摘要:#include <iostream.h>
#include <fstream.h>
ifstream f("e.in");
#define cin f
int main()
{
    int n,m;
    int a[10000];
    int b[5000];
    int i,j;
    int sum=0,total;
    while(cin>>n>>m)
    {
        for(i=0;i<m;i++)
        b[i]=0;
        sum=total=0;
        for(i=0;i<n;i++)
        cin>>a[i];
        for(i=0;i<n;i++)
        {
            sum+=a[i];
    ......

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