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 ;

}