(PAT 1065) A+B and C (64bit) (整数溢出问题)

柔情只为你懂 2022-03-19 15:46 226阅读 0赞

Given three integers A, B and C in \[−2^63,2^63\], 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, Band 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

解题思路:

整数相加溢出问题

如果两个正数相加结果<=0,则发生了溢出

如果两个负数相加结果>=0,则发生了溢出

如果正数发生了溢出,那么它们相加一定比结果要大,如果负数发生了溢出,那么它们相加一定比结果要小

分情况讨论即可

#define _CRT_SECURE_NO_WARNINGS
    #include <iostream>
    #include <algorithm>
    #include <string>
    using namespace std;
    
    int main() {
    	int N;
    	bool flag = false;
    	scanf("%d", &N);
    	long long a, b, c;
    	for (int i = 0; i < N; ++i) {
    		scanf("%lld %lld %lld", &a, &b, &c);
    		long long res = a + b;
    		
    		if (a > 0 && b > 0 && res <= 0) {
    			flag = true;
    		}
    		else if (a < 0 && b < 0 && res >= 0) {
    			flag = false;
    		}
    		else {
    			if (a + b > c) {
    				flag = true;
    			}
    			else {
    				flag = false;
    			}
    		}
    		if (flag) {
    			printf("Case #%d: true\n", i + 1);
    		}
    		else {
    			printf("Case #%d: false\n", i + 1);
    		}
    	}
    
    	system("PAUSE");
    	return 0;
    }