PAT A1065(20) A+B AND C

1065 A+B and C (64bit) (20)（20 分）提问

Given three integers A, B and C in [-2^63^, 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&gtC, 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

思路:

          题目给出的范围[-2^63,2^63] 应该用long long.

          隐含的条件就是发生了正溢出和负溢出,正溢结果为负数,负溢结果为正,需要考虑这一点.

       

#include<cstdio>
int main()
{
 
	int i,K;
	scanf("%d",&K);
	long long  a,b,c;  //长整数类型 
	bool jud;
	for(i=1;i<=K;i++)
	{
		scanf("%lld%lld%lld",&a,&b,&c);
		long long sum=a+b;			//二者加和值也必为长整数类型 
		if(a>0&&b>0&&sum<0) 		jud=true;		//a,b均为正数，且加值发生溢出，若溢出值为负数，则a+b>c; 
		else if(a<0&&b<0&&sum>=0)	jud=false;		//a,b均为负数，且加值发生溢出，若溢出值为正数，则a+b<c;
		else if(sum>c) 				jud=true;		//如果加值大于C，则说明没有溢出，且a+b>c; 
		else 						jud=false;		//其他情况，则为a+b<c; 
		if(jud==true)	printf("Case #%d: true\n", i);       //如果判断值为正，输出相应反应； 
		if(jud==false)	printf("Case #%d: false\n", i);      //如果判断值为负，输出相应反应； 
	}
	return 0;
 }