PAT A1065 A+B and C (64bit)-蒲公英云

PAT A1065 A+B and C (64bit)

桃扇骨 2021-10-01 05:54 311阅读 0赞

Given three integers A, B and C in [−2的63次方 ,2的63次方 ], 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

接下来时AC代码：

#include <stdio.h>
#include <stdbool.h>
int main(int argc, char *argv[]) { 
    int T,count=1;
    long long a,b,c;
    long long result;
    bool flag;
    scanf("%d",&T);
    while(T--){ 
        scanf("%lld%lld%lld",&a,&b,&c);
        result=a+b;
        if(a>0&&b>0&&result<0){ 
            flag=true;         //正溢出为true 
        }else if(a<0&&b<0&&result>=0){ 
            flag=false;        //负溢出为false 
        }else if(result>c){ 
            flag=true;         //无溢出时，正常比较 
        }else{ 
            flag=false;           //无溢出时，正常比较 
        }
        if(flag==true){ 
            printf("Case #%d: true\n",count++);
        }else{ 
            printf("Case #%d: false\n",count++);
        }
    }
    return 0;
}

注意：
1.这个题目的主要意图是让我们考虑溢出
2.计算机组成原理中指出，两个正数之和等于复数或者两个负数之和等于正数，就是溢出
3.a>0,b>0,a+b<0的时候为正溢出，结果为true 4.a<0,b<0,a+b>=0的时候为负溢出，结果为false
5.没有发生溢出时正常结算
6.a+b必须放到longlong类型变量中才可以和c进行比较，直接用a+b和c比较会造成数据错误