PAT A1065 A+B and C (64bit)

桃扇骨 2021-10-01 05:54 250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