A1065 A+B and C (64bit)

decription

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>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

idea

• A,B的范围为 [-2^63, 2^63]，
  则A, B, C,A+B都可能会超出long long 范围
  结合本题只有20分，不至于用到大数运算
• 考虑分类讨论，记sum = A+B
  当sum正溢出时，sum<0，则为true
  当sum负溢出时，sum>=0，则为false(注意等于的情况，否则第三个测试点过不去
  当sum为发生溢出时，直接比较sum和c的大小即可

solution1

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

solution2

#include <stdio.h>
int main(){
	int n;
	while(scanf("%d", &n) != EOF && n){
		for(int i = 1; i <= n; i++){
			long long a, b, c;
			scanf("%lld%lld%lld", &a, &b, &c);
			if(((a + b) > c && (a > 0 || b > 0)) || (a > 0 && b > 0 && (a + b) < 0))
				printf("Case #%d: true", i);
			else
				printf("Case #%d: false", i);
			if(i != n)
				printf("\n");
		}
	}
	return 0;
}