POJ Problem 2562 Primary Arithmetic
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11358 | Accepted: 4172 |
Description
Children are taught to add multi-digit numbers from right-to-left one digit at a time. Many find the "carry" operation - in which a 1 is carried from one digit position to be added to the next - to be a significant challenge. Your job is to count the number
of carry operations for each of a set of addition problems so that educators may assess their difficulty.
Input
Each line of input contains two unsigned integers less than 10 digits. The last line of input contains 0 0.
Output
For each line of input except the last you should compute and print the number of carry operations that would result from adding the two numbers, in the format shown below.
Sample Input
123 456 555 555 123 594 0 0
Sample Output
No carry operation. 3 carry operations. 1 carry operation.
Source
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>
#include <iostream>
#define MAX_N 106
#define MAX(a, b) (a > b)? a: b
#define MIN(a, b) (a < b)? a: b
using namespace std;
int main() {
int a, b;
int s1[MAX_N], s2[MAX_N];
while (scanf("%d%d", &a, &b) ,a || b) {
for (int i = 0; i < 20; i++) {
s1[i] = s2[i] = 0;
}
int ans = 0;
int cnt = 0;
while (a) {
s1[cnt++] = a%10;
a /= 10;
}
cnt = 0;
while (b) {
s2[cnt++] = b%10;
b /= 10;
}
for (int i = 0; i < 11; i++) {
s1[i] += s2[i];
if (s1[i] > 9) {
ans++;
s1[i + 1]++;
}
}
if (!ans) printf("No carry operation.\n", ans);
else if (ans == 1) printf("%d carry operation.\n", ans);
else printf("%d carry operations.\n", ans);
}
return 0;
}
