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
题意：
给出三个整数， 判断前两个的和是不是大于第三个。
分析：
考察溢出的检测问题, 当A和B同时为正整数或者同时为负整数的时候要小心题目的和可能溢出,
所以不能简单的return A+B > C, 要加溢出的判断代码。
#define LLONG_MAX 9223372036854775807i64
/* maximum signed long long int value */
#define LLONG_MIN (-9223372036854775807i64 - 1) /* minimum signed long long int value */
代码：
#include <iostream>
#include <fstream>
#include <climits>
using namespace std;
ifstream fin("in.txt");
#define cin fin
bool Compare(const long long& a,const long long& b,const long long& c){
if(a>0 && b>0){
if(a > LLONG_MAX - b)return true;
}
if(a<0 && b<0){
if(a < LLONG_MIN -b)return false;
}
return a+b > c;
}
int main()
{
int n;
cin>>n;
long long a,b,c;
int i;
for(i=0;i<n;i++)
{
cin>>a>>b>>c;
cout<<"Case #"<<i+1<<": ";
if(Compare(a,b,c)){
cout<<"true"<<endl;
}else
cout<<"false"<<endl;
}
system("PAUSE");
return 0;
}