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  • 【ATcoder】AtCoder Beginner Contest 163

    A - Circle Pond

    计算圆的面积

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    using namespace std;
    const double Pi = acos(-1);
    double r;
    int main() {
        cin >> r;
        printf("%.20lf", 2 * Pi * r);
        return 0;
    }

    B - Homework

    全部加起来,如果n小于和则输出-1,否则输出n-和。

    #include <iostream>
    #include <cstdio>
    using namespace std;
    int n, m;
    int a[10010], sum;
    int main() {
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= m; i++) scanf("%d", &a[i]), sum += a[i];
        if (n < sum) {
            puts("-1");
        } else {
            cout << n - sum;
        }
        return 0;
    }

    C - management

    输出出度即可

    #include <iostream>
    #include <cstdio>
    using namespace std;
    const int N = 400010;
    int n;
    int out[N];
    int main() {
        cin >> n;
        for (int i = 2; i <= n; i++) {
            int a;
            cin >> a;
            out[a]++;
        }
        for (int i = 1; i <= n; i++) {
            cout << out[i] << "
    ";
        }
        return 0;
    }

    D - Sum of Large Numbers

    题解待填坑

    #include <iostream>
    #include <cstdio>
    using namespace std;
    typedef long long ll;
    const ll mod = 1e9 + 7;
    const ll N = 200010;
    ll n, k, ans;
    ll presum[N], sufsum[N];
    int main() {
        cin >> n >> k;
        for (ll i = 0; i <= n; i++) {
            presum[i] = presum[i - 1] + i;
        }
        for (ll i = n; i >= 0; i--) {
            sufsum[i] = sufsum[i + 1] + i;
        }
        for (ll i = k; i <= n + 1; i++) {
            ans = (ans + (sufsum[n - i + 1] - presum[i - 1] + 1)) % mod;
        }
        cout << ans;
        return 0;
    }

    E - Active Infants

    待填坑

    F - path pass i

    待填坑

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  • 原文地址:https://www.cnblogs.com/zcr-blog/p/12790730.html
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