博文
stu(2147)(2005-08-30 20:21:00)
摘要：#include <iostream.h>
#include <string.h>
int fun(char s)
{
if (s=='V') return 0;
if (s=='U') return 1;
if (s=='C') return 2;
if (s=='D') return 3;
}
int main()
{
char a[4];
a[0]='V';a[1]='U';a[2]='C';a[3]='D';
char str1[9],str2[9],str3[9];
char op;
int i,j,n,k,first=0;
cin>>n;
for (k=0;k<n;k++)
{
cin>>str1>>str2;
first++;
int len=strlen(str1);
for (i=0;i<len;i++)
str1[7-i]=str1[len-i-1];
for (i=0;i<8-len;......
stu(1191)续(2005-08-30 20:21:00)
摘要：int main(void) {
char a, b, c, d;
int n, i;
char s[5];
scanf( "%d", &n);
for (i=0; i<n; ++i) {
scanf( "%s", s);
test( s[0], s[1], s[2], s[3]);
}
return 0;
}
void convert( char c, int i) {
switch (c) {
case 'A':
cardsnum[i] = 1;
break;
case '2':
cardsnum[i] = 2;
break;
case '3':
&nbs......
stu(1191)(2005-08-30 20:20:00)
摘要：#include <stdio.h>
#include <math.h>
#define TheBigNumber 9999999 //出错信息
#define Epslon 0.0000000001
typedef double (*pfddd)( double, double);
double add( double, double);
double sub( double, double);
double mult( double, double);
double div( double, double);
void convert( char, int);
int cardsnum[4];
int is24 (double x) { return fabs(x - 24) < Epslon ; }
void test( char a, char b, char c, char d) {
pfddd f[4] = {add, sub, mult, div};
char o[4] = {'+', '-', '*', '/'};
int i, j, k, l, m, n;
convert( a, 0);
convert( b, 1);
convert( c, 2);
convert( d, 3);
for (i = 0; i < 4; ++i)
for (j = 0; j <......
stu(1185)(2005-08-30 20:19:00)
摘要：# include<iostream.h>
int vexNum,arcNum;
long weight[100];
int arcs[100][100];
int dis[100][100];//dis[v][w]表示v到w的最少权值
long result[100][100];
int k;
long sum;
#define maxValue 32767
void init()
{
int i,j;
for(i=0;i<vexNum;i++)
cin>>weight[i];
for(i=0;i<100;i++)
for(j=0;j<100;j++){
if(i==j)
arcs[i][j]=0;
else
arcs[i][j]=maxValue;
}
......
stu(1162)(2005-08-30 20:18:00)
摘要：#include <stdio.h>
#include <math.h>
int change(int test); /* 判定函数 测试TEST */
int fun(int num); /* 判断NUM中可以找到最大的2^n 返回n */
int nn(int n); /* 返回2^n的大小 */
void main()
{
int n=0;
scanf("%d",&n);
while(n)
{
change(n);
printf("\n");
scanf("%d",&n);
......
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......
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......
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
}
......
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......
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;
......