题目
源地址:
http://poj.org/problem?id=2562
理解
高精度的模拟加法进位,数组模拟之。
代码
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <numeric>
#include <iomanip>
#include <bitset>
#include <sstream>
#include <fstream>
#define debug puts("-----")
#define pi (acos(-1.0))
#define eps (1e-8)
#define inf (1<<28)
using namespace std;
char num1[20];
char num2[20];
int main(int argc, char const *argv[])
{
int add, tmp;
int carry;
int i, j;
int len1, len2;
while (scanf("%s %s", num1, num2), strcmp(num1, "0") != 0 || strcmp(num2, "0") != 0){
getchar();
len1 = strlen(num1);
len2 = strlen(num2);
for (carry = add = 0, i = len1 - 1, j = len2 - 1 ; i >= 0 && j >= 0; i--, j--){
if ((num1[i] - '0') + (num2[j] - '0') + add >= 10){
add = 1;
carry ++;
}
else{
add = 0;
}
}
for (; i >= 0; i--){
if (num1[i] - '0' + add >= 10){
carry++;
add = 1;
}
else {
add = 0;
}
}
for (; j >= 0; j--){
if (num2[j] - '0' + add >= 10){
carry++;
add = 1;
}
else {
add = 0;
}
}
if (carry == 0){
printf("No carry operation.\n");
}
else if (carry == 1){
printf("1 carry operation.\n");
}
else{
printf("%d carry operations.\n", carry);
}
}
return 0;
}
更新日志
- 2014年08月22日 已AC。