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
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#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; } |
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