A1065. A+B and C (64bit)

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

 #include<cstdio>
 #include<iostream>
 using namespace std;
 int main(){
     int T;
     long long a, b, c, test;
     scanf("%d", &T);
     for(int i = ; i < T; i++){
         scanf("%lld%lld%lld", &a, &b, &c);
         test = a + b;
         if(a > && b >  && test < ){
             printf("Case #%d: true\n", i + );
         }else if(a <  && b <  && test >= ){
             printf("Case #%d: false\n", i + );
         }else{
             if(test > c)
                 printf("Case #%d: true\n", i + );
             else
                 printf("Case #%d: false\n", i + );
         }
     }
     return ;
 }

总结：1、首先输入的范围在[-2^63, 2^63], 则只能用long long 8个byte存储。当a与b过大时，会产生溢出，需要考虑正数+正数=负数及负数+负数=正数的情况。

　　　2、A + B的结果不能在条件语句中直接与C比较，必须先存储于long long变量中。