【高精乘单精】岛屿的缘分

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思路：

裸的高精乘单精

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$Code$Code:

#include <cstdio>
#include <iostream>
#define CWH using
#define AK namespace
#define IOI std
CWH AK IOI;
char x[500],y[500];
int k,a[500],b[500],ans[500],g;
int main ()
{
	//freopen("b.in","r",stdin);freopen("b.out","w",stdout);
	scanf("%s%s%d",x,y,&k);
	int len1 = strlen(x),len2 = strlen(y);
	for (int i = 0; i < len1; ++i) a[len1 - i] = x[i] - '0'; //倒置
	for (int i = 0; i < len2; ++i) b[len2 - i] = y[i] - '0';
	for (int i = 1; i <= len1; ++i)
	{
		 g = 0;
	    for (int j = 1; j <= len2; ++j)
	    {
	  	  ans[i + j - 1] = ans[i + j - 1] + a[i] * b[j] + g;
		  g = ans[i + j - 1] / 10;
		  ans[i + j - 1] %= 10;//高精乘
	    }
	  ans[i + len2] = g;
    }
    for (int i = k; i >= 1; --i)
    printf("%d",ans[i]);//倒叙输出
}