PAT--1065. A+B and C (64bit)
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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
题解
考虑到a, b, c
的取值范围,这题并不需要模拟大数的加法。
有符号long long
的范围是 -
a + b
只有同为正数或同为负数才有可能溢出,溢出必然变号,也肯定大于(小于)c
。
#include <bits/stdc++.h>using namespace std;long long a, b, c;int n;int main(){ cin >> n; for(int i = 1; i <= n; ++i){ cin >> a >> b >> c; cout << "Case #" << i << ": "; long long t = a + b; if(a > 0 && b > 0 && t < 0) cout << "true" << endl; else if(a < 0 && b < 0 && t >= 0) cout << "false" << endl; else if(t > c) cout << "true" << endl; else cout << "false" << endl; } return 0;}
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