1065. A+B and C (64bit) (20)-PAT甲级真题

来源：互联网发布：pano2vr mac v4.1.0 编辑：程序博客网时间：2024/06/05 16:39

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;}

0 0