A1065.A+B and C (64bit) (20)-优快云博客

本文介绍了一种通过给定三个整数A、B和C，判断A+B是否大于C的方法。提供了两种不同的C语言实现方案，一种利用了整数溢出的特性，另一种则避免了直接使用可能溢出的操作。

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

参考代码1：这里没有用到溢出的知识

#include<cstdio>
const int T=10;
int main(){
	long long a[T],b[T],c[T];
	int t;
	scanf("%d",&t);
	for(int i=0;i<t;i++){
		scanf("%lld %lld %lld",&a[i],&b[i],&c[i]);
	}
	for(int i=0;i<t;i++){
		if(a[i]>0&&b[i]>0){
			if(a[i]>9223372036854775808-b[i]) printf("Case #%d: true\n",i+1);
			else if(a[i]+b[i]>c[i]) printf("Case #%d: true\n",i+1);
			else printf("Case #%d: false\n",i+1);
		}else if(a[i]<0&&b[i]<0){
			if(a[i]<-9223372036854775808-b[i]) printf("Case #%d: false\n",i+1);
			else if(a[i]+b[i]>c[i]) printf("Case #%d: true\n",i+1);
			else printf("Case #%d: false\n",i+1);
		}else{
			if(a[i]+b[i]>c[i]) printf("Case #%d: true\n",i+1);
			else printf("Case #%d: false\n",i+1);
		}
	}
	return 0;
}

参考代码2：用溢出相关的知识可以得到更简洁的代码

#include<cstdio>
int main(){
	long long a,b,c;
	int t;
	int i=1;
	scanf("%d",&t);
	while(t--){
		scanf("%lld %lld %lld",&a,&b,&c);
		long long ans=a+b;
		bool flag;
		if(a>0&&b>0&&ans<=0) flag=true;
		else if(a<0&&b<0&&ans>=0) flag=false;
		else if(ans>c) flag=true;
		else flag=false;
		if (flag==true) printf("Case #%d: true\n",i);
		else printf("Case #%d: false\n",i);
		i++;
	}
	return 0;
}

转载于:https://www.cnblogs.com/cobread/p/10846256.html