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

---

判断a+b是否大于c。这里需要注意的是a+b可能溢出，如果a大于0且b大于0，但相加得到的却是小于等于0的说明是正溢出，这时肯定比c大（因为c肯定在long long的范围内）。如果a小于0且b小于0，但相加得到的却是大于等于0的说明是负溢出，这是肯定比c小。其他情况就和平常计算一样。这里还要注意a+b要赋值给一个变量再和c比较。

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;

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