poj 上的1006题runtime error，高手帮忙看下
用的是余数定理做的，在自己机子上感觉速度蛮快，为什么会runtime error？
代码风格不好，看着可能有点累，麻烦大家帮忙看下

C/C++ code

#include<stdio.h>#include<iostream>#define F 23#define S 28#define T 33#define MAX 50int main(){    int f,s,t,d,a[MAX][4],min,i=0,j=0,k1,k2,k3,number;    for(scanf("%d %d %d %d",&f,&s,&t,&d);(f!=-1 && s!=-1 && t!=-1 && d!=-1);scanf("%d %d %d %d",&f,&s,&t,&d))         {           //将四个输入的值存入数组中，如果输入为-1，-1，-1，-1结束 ，                       a[i][j]=f;                   a[i][j+1]=s;                   a[i][j+2]=t;                   a[i][j+3]=d;                   i++;         }         min=F*S*T;//最小公倍数          number=i;//输入数的数目          for(i=0;i<number;i++)//用剩余定理计算         {             int sum=0,temp1,temp2,temp3;             k1=temp1=min/F;             k2=temp2=min/S;             k3=temp3=min/T;             int m1=1,m2=1,m3=1;             while((k1%F)!=1){                   ++m1;                   k1=temp1*m1;      //k1、k2、k3存储用于计算的关键值键值                   }             while((k2%S)!=1){                           ++m2;                           k2=temp2*m2;                           }             while((k3%T)!=1){                           ++m3;                           k3=temp3*m3;                           }             sum=k1*a[i][j]+k2*a[i][j+1]+k3*a[i][j+2];                                       sum=sum%min-a[i][j+3];             if(sum<=0)                sum=sum+min;             printf("Case %d: the next triple peak occurs in %d days.\n",i+1,sum);         }         system("PAUSE");         return 0;}

C/C++ code

#include<iostream>//#include<vector>using std::cin;using std::cout;using std::endl;//using std::vector;int main(int argc, char *argv[]){    int p = 0, e = 0, i = 0, d = 0;        int cases = 0;    while(cin >> p >> e >> i >> d)    {        if(-1 == p)            break;                ++cases;                int j = 1;        while(j <= 21252)        {            if((j + d - p) % 23 ==0 && (j + d - e) % 28 ==0 && (j + d - i) % 33 ==0)            {                cout << "Case " << cases << ": the next triple peak occurs in " << j << " days." << endl;                break;            }            ++j;        }    }    return 0;}
[解决办法]
探讨

问题解决了，不能将每组输入的数用数组存储
要每输入一组数就输出结果
同时谢谢楼上的朋友

[解决办法]
做了一下，解决了：代码如下：[code=C/C++][/code]#include<iostream>
using namespace std;
int main()
{int p,e,i,d,a,t=0;
while(1)
{
scanf("%d%d%d%d",&p,&e,&i,&d);
if(p==-1 && e==-1 && i==-1 && d==-1)
break;
a=(5544*p+14421*e+1288*i-d+21252)%21252;
if(!a)
a=21252;
printf("Case %d: the next triple peak occurs in %d days.\n",++t,a);
}
return 0;
}