Pat--a+b and C (study type data storage)

A+b and C (15) time limit MS Memory limit 32768 KB code length limit of KB Judging program standard (from small) Title Description

The 3 integers a, B, and C in a given interval [2 of 31, 2 of 31], determine if A+B is greater than C.

Input Description:

Enter line 1th to give the positive integer t (<=10), which is the number of test cases. The T-group test cases are then given, one row for each group, and a, B, and C in order. The integers are separated by a space.

Output Description:

For each set of test cases, output "case #X: True" in one row if a+b>c, otherwise the output is "case #X: false", where X is the number of the test cases (starting at 1).

Input Example:

4

1 2 3

2 3 4

2147483647 0 2147483646

0-2147483648-2147483647

Output Example:

 case #1: false 
 
 Case #2: True 
 
 Case #3: True 
 
 Case #4: false 
 
  This problem is mainly not directly to compare the size of a+b and C, need to divide the situation, the indirect to discuss 
 below is the code of AC 
  

#include <iostream> #define LINT Long Long intusing namespace Std;int main () {LINT n;cin>>n; LINT K=1;while (n--) {LINT a,b,c;cin>>a>>b>>c;if ((a>=0&&b<=0) | | (a<=0&&b>=0)) {if (a+b>c) cout<< "Case #" <<k<< ": True" <<endl;elsecout<< "Case #" <<k<< " : false "<<endl;k++;} else if (a>0&&b>0) {if (c<=a| | C<=B) cout<< "Case #" <<k<< ": True" <<endl;else{if (c-a>=b) cout<< ' Case # ' <<k << ": false" <<endl;elsecout<< "Case #" <<k<< ": true" <<ENDL;} k++;} else if (a<0&&b<0) {if (c>=a| | C>=B) cout<< "Case #" <<k<< ": false" <<endl;else{if (c-a>=b) cout<< ' Case # ' << k<< ": false" <<endl;elsecout<< "Case #" <<k<< ": true" <<ENDL;} k++;}}}

　　

Pat--a+b and C (study type data storage)