NYOJ 棋盘覆盖

数字很大，要用大数乘法。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
char s1[];
char s2[];
char* bignum(char *num1, char *num2)
{
    if(strcmp(num1,"")==)
        return "";
    int length1 = strlen(num1);
    int length2 = strlen(num2);
    int i, l;
    char *res = (char *)malloc(sizeof(char)*(length1 + length2)); //开辟相应内存
    memset(res, , sizeof(char)*(length1 + length2));
    for (i = length1 - ; i >= ; i--)
    for (l = length2 - ; l >= ; l--)
    {
        res[i + l + ] += (num1[i] - '')*(num2[l] - '');
        res[i + l] += res[i + l + ] / ;    //马上进行进位
        res[i + l + ] %= ;
    }
    int count = ;
    while (res[count] == )  //由于保存的结果是从右向左的，所以要消除左部分的0；
    {
        count++;
    }
    char* ret = (char *)malloc(sizeof(char)*(length1 + length2 + ));
    memset(ret, , sizeof(char)*(length1 + length2 + ));
    for (l = , i = count; i < length1 + length2; l++, i++)  //非0部分赋给ret
    {
        ret[l] = res[i] + '';
    }
    //printf("Ret=%s\n", ret);
    return ret;
    free(res);
    free(ret);
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int k;
        cin>>k;
        strcpy(s1,"");
        strcpy(s2,"");
        //printf("%s\n",s1);
        while(k--)
        {
            //printf("%s\n",s1);
            strcpy(s1,bignum(s1,s2));
            for(int i=strlen(s1)-;i>=;i--)
            {
                if(s1[i] == '')
                {
                    s1[i] = ;
                    continue;
                }
                else
                {
                    s1[i] = s1[i] + ;
                    break;
                }
            }
        }
        printf("%s\n",s1);
    }
    return ;
}