sicily 1020 Big Integer

 大数取模，把输入的大数的字符串转换为整数，但同时每一步要取模（老是错在这里），不然就爆了int！
 ----> temp = ((temp * 10) + (s[i]-'0')) % a[j];
 1 #include <iostream>
#include <cstdio>
#include <cstring>
#include <stack>

using namespace std;

int a[105];
char s[505];

int main()
{
    int t;
    cin >> t;
    while(t--)
    {
    
        int n;
        cin >> n;
        if(n == 0)
        {
            cout << "(" << 0 << ")" << endl;
            continue;
        }
        for(int i=0; i<n; i++)
            cin >> a[i];
    
        cin >> s;
        int len = strlen(s);
        
        cout << "(";
        for(int j=0; j<n; j++)
        {
            int temp=0;
            for(int i=0; i<len; i++)
                temp = ((temp * 10) + (s[i]-'0')) % a[j]; //每一步都要取模 
            
            if(n == 1)
                cout << temp%a[j] << endl;
            else
            {
                if(j == 0)
                {
                    cout << temp%a[j];
                    continue;
                }
                    
                cout << "," << temp%a[j];
                
            }
        }
        cout << ")" << endl;
    
        
    }
    return 0;
}