PAT A1065 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

题意:

判断a+b是否大于c，是输出Case #X: true，否输出Case #X: false；

输入：

测试样例数
a b c
. . .

输出：

Case #1: false（或true）
. . .

思路：

（1）a，b，c的范围为[-2⁶³,2⁶³）,确定其数据类型为long long型；
（2）要注意a+b的临界情况，若a>0,b>0,a+b<=0,则正溢出（参考计算机组成原理），输出Case #X: true;若a<0,b<0,a+b>=0,则负溢出，输出Case #X： false；
（3）在 if 语句的（）中将long long 型的a+b直接与c比较会出错，所以需要用temp来存放a+b.

代码：

#include <cstdio>
int main(){
	long long a,b,c,temp;
	int t;
	scanf("%d",&t);
	for(int i=1;i<=t;i++){
		scanf("%lld %lld %lld",&a,&b,&c);
		temp=a+b;
		if(a>0&&b>0&&temp<=0){
			printf("Case #%d: true\n",i);
		}else if(a<0&&b<0&&temp>=0){
			printf("Case #%d: false\n",i);
		}else if(temp>c){
			printf("Case #%d: true\n",i);
		}else if(temp<=c){
			printf("Case #%d: false\n",i);
		}
	}
	return 0;
}

词汇：

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