PAT甲级 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

想用cin必须得每次输入clear一下,否则由于溢出的问题,输入数据会存在问题

#include <iostream>


using namespace std;
typedef long long LL;

int main() {
    int N;
    LL a, b, c;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        cin >> a;
        cin.clear();
        cin >> b;
        cin.clear();
        cin >> c;
        cin.clear();
        LL sum = a + b;
        bool flag = false;
        if (a > 0 and b > 0 and sum < 0)
            flag = true;
        else if (a < 0 and b < 0 and sum >= 0)
            flag = false;
        else if (sum > c)
            flag = true;

        if (flag)
            cout << "Case #" << i + 1 << ": true" << endl;
        else
            cout << "Case #" << i + 1 << ": false" << endl;
    }
    return 0;
}

当然直接scanf就可以解忧愁了

#include <iostream>


using namespace std;
typedef long long LL;

int main() {
    int N;
    LL a, b, c;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        scanf("%lld %lld %lld", &a, &b, &c);
        LL sum = a + b;
        bool flag = false;
        if (a > 0 and b > 0 and sum < 0)
            flag = true;
        else if (a < 0 and b < 0 and sum >= 0)
            flag = false;
        else if (sum > c)
            flag = true;
        if (flag)
            cout << "Case #" << i + 1 << ": true" << endl;
        else
            cout << "Case #" << i + 1 << ": false" << endl;
    }
    return 0;
}
#include <iostream>

using namespace std;

long double a, b, c;

int main()
{
    int T;
    cin >> T;
    for (int i = 1; i <= T; i ++ )
    {
        printf("Case #%d: ", i);
        cin >> a >> b >> c;
        if (a + b > c) puts("true");
        else    puts("false");
    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包

打赏作者

XdpCs

你的鼓励将是我创作的最大动力

¥1 ¥2 ¥4 ¥6 ¥10 ¥20
扫码支付:¥1
获取中
扫码支付

您的余额不足,请更换扫码支付或充值

打赏作者

实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值