前言

传送门

正文

题目描述

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

思路：

由于long long的范围为[-2-63,263-1],故两数相加可能会溢出(正溢出/负溢出)，当a,b均大于0，而a+b<0，此时为正溢出；当a,b均小于0，而a+b>=0，此时为负溢出。而易知，正溢出时，a+b必定大于c，而负溢出时a+b必定小于c

参考题解:

/*
注意正溢出和负溢出的问题 
*/
#include<cstdio>
#define LL long long
int main(){
	LL a,b,c,sum;
	int t,count=1;
	scanf("%d",&t);
	bool flag;
	while(t--){
		scanf("%lld%lld%lld",&a,&b,&c);
		sum=a+b;
		if(a>0&&b>0&&sum<0)flag=true;
		else if(a<0&&b<0&&sum>=0)flag=false;
		else if(a+b>c)flag=true;
		else flag=false;
		if(flag==true){
			printf("Case #%d: true\n",count++);
		}else{
			printf("Case #%d: false\n",count++);
		}
	}
	return 0;
}

后记

穿越人海，只为与你相拥；
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