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[1000];
char s2[10];
char* bignum(char *num1, char *num2)
{
    if(strcmp(num1,"0")==0)
        return "0";
    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, 0, sizeof(char)*(length1 + length2));
    for (i = length1 - 1; i >= 0; i--)
    for (l = length2 - 1; l >= 0; l--)
    {
        res[i + l + 1] += (num1[i] - '0')*(num2[l] - '0');
        res[i + l] += res[i + l + 1] / 10;    //马上进行进位
        res[i + l + 1] %= 10;
    }
    int count = 0;
    while (res[count] == 0)  //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 + 2));
    memset(ret, 0, sizeof(char)*(length1 + length2 + 2));
    for (l = 0, i = count; i < length1 + length2; l++, i++)  //非0部分赋给ret
    {
        ret[l] = res[i] + '0';
    }
    //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,"0");
        strcpy(s2,"4");
        //printf("%s\n",s1);
        while(k--)
        {
            //printf("%s\n",s1);
            strcpy(s1,bignum(s1,s2));
            for(int i=strlen(s1)-1;i>=0;i--)
            {
                if(s1[i] == '9')
                {
                    s1[i] = 0;
                    continue;
                }
                else
                {
                    s1[i] = s1[i] + 1;
                    break;
                }
            }
        }
        printf("%s\n",s1);
    }
    return 0;
}