高精度*高精度
思路来源:
http://blog.youkuaiyun.com/luojj26/article/details/49671121
http://wenku.baidu.com/view/233cb73331126edb6f1a1030.html
#include "stdafx.h"
#include<stdio.h>
#include<string.h>
int main() {
char numberN[1500], numberM[1500];
scanf("%s%s", numberN, numberM);
int n = strlen(numberN), m = strlen(numberM);
int a[3], b[2];
int i, j;
for (i = 0, j = n - 1; i < n; i++, j--) {
a[i] = numberN[j] - '0';
}
for (i = 0, j = m - 1; i < m; i++, j--) {
b[i] = numberM[j] - '0';
}
int c[3000];
for (i = 0; i < 3000; i++) {
c[i] = 0;
}
for (i = 0; i < n; i++) {
for (j = 0; j < m; j++) {
c[i + j] += a[i] * b[j];
}
}
for (i = 0; i < n + m; i++) {
if (c[i] >= 10) {
c[i + 1] += c[i] / 10;
c[i] %= 10;
}
}
/*
for(i = 0; i < 3; i++)
{
c[i] = (a[i] + b[i]) % 10 + c[i]; //对最低位2个数相加,得到2位数的个位数值。
c[i+1] = (a[i] + b[i])/10; //对最低位2个数相加,若大于10,进位。
}
*/
for (j = 2999; j > 0; j--) {
if (c[j] != 0)
break;
}
for (i = j; i >= 0; i--) {
printf("%d", c[i]);
}
printf("\n");
return 0;
}