PAT (Advanced Level) Practice 1065

1065 A+B and C (64bit) (20)（20 point(s)）

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

题意，输入A和B和C（都在long范围内），请问A+B是否C。

思路：1.要么写个大数，直接相加。

          2.分情况讨论。

            (1) A和B都小于0，A和B都大于0情况，那么只需要通过C-A和B进行比较就能判断。

            (2)A和B异号，那么直接相加即可，不存在溢出。

代码：

#include
#include
typedef long long ll;
int main(){
	int n;
	scanf("%d", &n);
	for(int i = 0;i < n;i++){
		ll A, B, C;
		scanf("%lld%lld%lld", &A, &B, &C);
		if(A > 0 && B > 0){
			if(A > C || B > C){
				printf("Case #%d: true", i+1);//A和B有一个大于C，那么A+B肯定大于C 
			}else{ 
				ll result = C - A; //否则，A+B可能大于C也可能小于C，但是存在溢出，于是使用C-A的方式，肯定不会溢出 
				if(B > result){
					printf("Case #%d: true", i+1);
				}else{
					printf("Case #%d: false", i+1);
				}
			}
		}else if(A < 0 && B < 0){//同上 
			if(A < C || B < C){
				printf("Case #%d: false", i+1);
			} else {
				ll result = C - A;
				if(B > result){
					printf("Case #%d: true", i+1);
				}else{
					printf("Case #%d: false", i+1);
				}
			}
		}else{//第三种就是正负相加即可，肯定不存在溢出 
			ll result = A + B;
			if(result > C){
				printf("Case #%d: true", i+1);
			} else {
				printf("Case #%d: false", i+1);
			}
		}
		printf("\n")
	}
	return 0;
}