A1065 A+B and C (64bit) (20 point(s))

两数相加出现正溢出和负溢出

1. 原文

原题

Given three integers A, B and C in [−2⁶³,2^63​], 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, 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

2. 解析

给出三个整数A，B，C，判断A+B>C

由于long long范围为 [−2⁶³,2⁶³]，计算机组成中指出，两个正数相加为负数，两个负数相加为正数，则为溢出。

• A,B最大为 2⁶³-1，两数相加为2⁶⁴-2，超过long long正向最大值而出现正溢出，正溢出区间[-2⁶³,-2]<0，

  (2⁶⁴-2)%2⁶⁴ = -2

  return true

• A,B最小为 -2⁶³，两数相加为-2⁶⁴，超过long long负向最小值而出现负溢出，负溢出区间[0,-2⁶³）>=0，

  -2⁶⁴%2⁶⁴ = 0

  return false

3. AC代码

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