PAT 1065. A+B and C (64bit) (20)

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:

31 2 32 3 49223372036854775807 -9223372036854775808 0

Sample Output:

Case #1: falseCase #2: trueCase #3: false思路：由于数字的范围，longlong还是差一点，这样的话通过除3稍微缩小范围可以判断。
#include <iostream>  using namespace std;  int main()  {  long long a, b, c;  int n;  cin >> n;  for(int i = 1; i <= n; i++){  cin >> a >> b >> c;  long long a1 = a / 3, b1 = b / 3, c1 = c / 3;  //a,b,c除于3，缩小范围，但注意处理余数long long a2 = a % 3, b2 = b % 3, c2 = c % 3;  long long sum1 = a1 + b1 + (a2 + b2) / 3, sum2 = (a2 + b2) % 3;//sum1是(a+b)/3的结果，注意余数部分可能有取整，sum2是(a+b)%3if(sum1 > c1 || (sum1 == c1 && sum2 > c2))  //(a+b)/3 > c/3 或者整数部分相等，余数比较大cout << "Case #" << i << ": true" << endl;  else  cout << "Case #" << i << ": false" << endl;  }  return 0;  }