1065 A+B and C (64bit) (20 分)
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<iostream>
using namespace std;
int main(){
int T;
cin>>T;
long long a,b,c;
for(int i=0;i<T;i++){
cin>>a>>b>>c;
long long sum = a+b;
if(a > 0 && b > 0 && sum < 0)
cout<<"Case #"<<i+1<<": true"<<endl;
else if(a < 0 && b < 0 && sum >= 0)
cout<<"Case #"<<i+1<<": false"<<endl;
else if(sum > c)
cout<<"Case #"<<i+1<<": true"<<endl;
else if(sum <=c)
cout<<"Case #"<<i+1<<": false"<<endl;
}
}