1065 A+B and C (64bit) (20分)

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

题目描述：A+B>C，真返回true，假返回false

解题思路：一看就是大数据处理，long double类型杠，或者对数据处理一下

注：a>>1 与 a/2有区别，(a&1) 与 a%2有区别，a+b==c 与 (g=a+b),g==c有区别

方法一：

#include<bits/stdc++.h>
using namespace std ;
int main()
{
    int n;
    scanf("%d", &n);
    for( int i=1; i<=n; i++){
        long double a, b, c;
        scanf("%Lf %Lf %Lf", &a, &b, &c);
        printf("Case #%d: ", i);
        if(a+b>c) printf("yes\n");
        else printf("false\n");
    }
    return 0 ;
}

方法二：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	long long a, b, c;
	int n;
	scanf("%d", &n);
	for(int i=1; i<=n; i++){
		scanf("%lld %lld %lld", &a, &b, &c);
		long long ta, tb, tc;
		ta = a>>1;
		tb = b>>1;
		tc = c>>1;
		printf("Case #%d: ", i);
		if(ta+tb>tc){
			printf("true\n");
		}
		else if(ta+tb==tc&&((a&1)+(b&1)>(c&1))){
			printf("true\n");
		}
		else printf("false\n");
	}
	return 0;
}

方法三：

n = int(input())
for i in range(1,n+1):
    g = input().split()
    a = int(g[0])
    b = int(g[1])
    c = int(g[2])
    if a+b > c:
        print('Case #{}: true'.format(i))
    else:
        print('Case #{}: false'.format(i))