【详细讲解】1065 A+B and C (64bit) (20 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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
1
2
3
4

Sample Output:
Case #1: false
Case #2: true
Case #3: false

数据溢出为环形，即263+1=2−632^{63}+1 = 2^{-63}263+1=2−63。同理：2−63−1=2632^{-63}-1= 2^{63}2−63−1=263

因此考虑两种情况：
溢出：如果a>0, b>0, 但a+b<0，即为溢出。溢出的值肯定比C大。反之同理。
不溢出：正常判断即可。

#include<bits/stdc++.h>
using namespace std;
using gg = long long;
int main() {gg n; scanf("%lld", &n); for(gg i = 1; i <= n; i++) {gg a, b, c;scanf("%lld %lld %lld", &a, &b, &c);gg res = a+b, flag;if(a>0 && b>0 && res<0) flag=true;        //都溢出了肯定比C大else if(a<0 && b<0 && res>=0) flag=false;//都溢出了肯定比C小else flag = a+b>c;printf("Case #%lld: %s\n", i, (flag ? "true" : "false")); } return 0;
}

耗时：

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【详细讲解】1065 A+B and C (64bit) (20 分)

立志用最少的代码做最高效的表达

耗时：

求赞哦~ (✪ω✪)

【详细讲解】1065 A+B and C (64bit) (20 分)相关推荐

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