| Product |
The Problem
The problem is to multiply two integers X, Y. (0<=X,Y<10250)
The Input
The input will consist of a set of pairs of lines. Each line in pair contains one multiplyer.
The Output
For each input pair of lines the output line should consist one integer the product.
Sample Input
12
12
2
222222222222222222222222
Sample Output
144
444444444444444444444444 题意: 大数相乘
做法, 看AC代码, 有注释, 自行理解, 不难~ AC代码:
#include<stdio.h>
#include<string.h> int main() {
char mul1[300], mul2[300];
char ch;
int ans[600];
int len1, len2;
int num1, num2;
int k, Sum;
while(gets(mul1) != NULL && gets(mul2) != NULL) {
memset(ans, 0, sizeof(ans));
len1 = strlen(mul1);
len2 = strlen(mul2);
//倒序第一个乘数:
for(int i = 0; i < len1/2; i++) {
ch = mul1[i];
mul1[i] = mul1[len1 - 1 - i];
mul1[len1 - 1 - i] = ch;
}
//倒序第二个乘数:
for(int i = 0; i < len2/2; i++) {
ch = mul2[i];
mul2[i] = mul2[len2 - 1 - i];
mul2[len2 - 1 - i] = ch;
}
//开始处理
for(int i = 0; i < len1; i++) {
num1 = mul1[i] - '0';
k = i;
for(int j = 0; j < len2; j++) {
num2 = mul2[j] - '0';
Sum = num1 * num2;
ans[k] = Sum % 10 + ans[k];
ans[k+1] = Sum / 10 + ans[k+1];
if(ans[k] > 9) {
ans[k+1] = ans[k] / 10 + ans[k+1];
ans[k] = ans[k] % 10;
}
k++;
}
}
//从后面开始查找第一个非零数, 然后倒序输出~
int pos;
for(int i = 599; i >= 0; i--) {
if(ans[i] != 0) {
pos = i;
for(int i = pos; i >= 0; i--)
printf("%d", ans[i]);
printf("\n");
break;
}
else if(i == 0 && ans[i] == 0)
printf("0\n");
}
}
return 0;
}