1065. A+B and C (64bit) (20)

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

---

• 笔记

long long 的范围是 [-2^63 , 2^63]
根据题目可知，直接进行a+b判断可能会发生溢出
两个正数之和等于负数 或者 两个负数之和等于正数 表示溢出
对溢出的结果进行判断可以确定 a+b 和 c 的大小

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