1065. A+B and C (64bit) (20)-PAT甲级真题

1065.
A+B and C (64bit) (20)

Given
three integers A, B and C in [-263, 263], you are supposed to tell whether A+B > C.

Input
Specification:

The
first line of the input gives the positive number of test cases, T (<=10). Then T test cases follow, each consists of a single line containing three integers A, B and C, separated by single spaces.

Output
Specification:

For
each test case, output in one line “Case #X: true” if A+B>C, or “Case #X: false” otherwise, where X is the case number (starting from 1).”

Sample
Input:

3

1
2 3

2
3 4

9223372036854775807
-9223372036854775808 0

Sample
Output:

Case
#1: false

Case
#2: true

Case
#3: false

分析：

因为A、B的大小为[-2^63,
2^63]，用long long 存储A和B的值，以及他们相加的值sum：

如果A
> 0, B < 0 或者 A < 0, B > 0，sum是不可能溢出的

如果A
> 0, B > 0，sum可能会溢出，sum范围理应为(0, 2^64 – 2]，溢出得到的结果应该是[-2^63, -2]是个负数，所以sum < 0时候说明溢出了

如果A
< 0, B < 0，sum可能会溢出，同理，sum溢出后结果是大于0的，所以sum > 0 说明溢出了

#include <cstdio>
using namespace std;
int main() {
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++) {
long long a, b, c;
scanf("%lld %lld %lld", &a, &b, &c);
long long sum = a + b;
if(a > 0 && b > 0 && sum < 0) {
printf("Case #%d: true\n", i + 1);
} else if(a < 0 && b < 0 && sum >= 0){
printf("Case #%d: false\n", i + 1);
} else if(sum > c) {
printf("Case #%d: true\n", i + 1);
} else {
printf("Case #%d: false\n", i + 1);
}
}
return 0;
}