UVA - 10106 Product 第二种解法

来源：互联网发布：linux下退出vi 编辑：程序博客网时间：2024/06/06 07:46

题目大意：求两个数的成积

解题思路：分步解决，一个个相乘再相加

#include<cstdio>#include<cstring>void add(char *sum, char * num2,int len) {int i;for( i = 0 ; i < len; i++) {int num = sum[i] + num2[i] - '0';sum[i] = num % 10;if(num > 9)sum[i+1] = sum[i+1] + 1;}}int main() {char num1[1000];char num2[1000];while( gets(num1)) {gets(num2);int len1 = strlen(num1);int len2 = strlen(num2);char temp[1000];char sum[2000] = {0};int j , l;for(int i = len1 - 1, k = 0; i >= 0; i--, k++)  {int mark = 0;for( j = len2 - 1, l = k; j >= 0; j--,l++)  {int num =( num1[i] - '0') * (num2[j] - '0') + mark;mark = num / 10;num = num % 10;temp[l] = num + '0';}  if(mark > 0) {temp[l] = mark + '0';l++;}add(sum,temp,l);memset(temp,'0',sizeof(temp));}int len3 = 1999;if(num1[0] == '0' && len1 == 1 || num2[0] == '0' && len2 == 1)printf("0");else {for(;;len3--)if(sum[len3] != 0)break;for(int i = len3; i >= 0; i--) printf("%d",sum[i]);}printf("\n");}return 0;}

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