Educational Codeforces Round 15 [111110]

注意一个词：连续

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<bits/stdc++.h>
using namespace std;
long long a[];
int main()
{
    //freopen("input.txt","r",stdin);
    int n;
    scanf("%d", &n);
    for (int i = ; i <= n; i++)
        scanf("%I64d", &a[i]);
    int ans = ;
    int i = , tmp = ;
    while (i <= n)
    {
        if (a[i - ] < a[i]) tmp++;
        else
        {
            if (tmp > ans) ans = tmp;
            tmp = ;
        }
        i++;
    }
    if (tmp > ans) ans = tmp;
    printf("%d\n", ans);
    //fclose(stdin);
    return ;
}

枚举每个可能的和，用map记下每种值有多少个，n个数过一遍，每次加上和减去当前值的差的值的个数。当当前值本身是2的幂时要额外减一。

int和longlong一起运算时最好把int弄成longlong

#include <iostream>
#include <map>
using namespace std;

map<long long, int> e;
long long a[];

int main()
{
    int n;
    cin >> n;
    for (int i = ; i <= n; i++)
        cin >> a[i];
    e.clear();
    for (int i = ; i <= n; i++)
        e[a[i]]++;
    long long sum = ;
    for (int i = ; i <= n; i++)
        sum += a[i];
    long long ans = ;
    for (int i = ; i <= n; i++)
    {
        long long tmp = ;
        for (long long j = ; j <= sum; j <<= )
        {
            if (j <= a[i]) continue;
            long long f = j - a[i];
            tmp += e[f];
            if (f == a[i]) tmp--;
        }
        ans += tmp;
    }
    cout << ans /  << endl;
    //system("pause");
    return ;
}

用二分法找到每个点最近的塔的距离，所有距离中最大的就是最小的满足条件的r。

#include <iostream>
#include <map>
using namespace std;

__int64 a[], b[];

int main()
{
    int n, m;
    cin >> n >> m;

    for(int i = ; i <= n; i++)
        cin >> a[i];

    for(int i = ; i <= m; i++)
        cin >> b[i];

    __int64 MAX = ;

    for(int i = ; i <= n; i++)
    {
        __int64 tmp = 1e15;

        if(a[i] <= b[]) tmp = b[] - a[i];
        else if(b[m] <= a[i]) tmp = a[i] - b[m];
        else
        {
            int l = , r = m, mid;

            while()
            {
                mid = (l + r) >> ;

                if(a[i] < b[mid]) r = mid;
                else l = mid;

                if(l == r || l +  == r) break;
            }

            tmp = a[i] - b[l];

            if(b[r] - a[i] < tmp) tmp = b[r] - a[i];
        }

        if(tmp > MAX) MAX = tmp;
    }

    cout << MAX << endl;
    //system("pause");
    return ;
}

跟VJ上的守望者的逃离差不多。

#include <iostream>
#include <map>
using namespace std;

int main()
{
    long long d, k, a, b, t;
    cin >> d >> k >> a >> b >> t;

    if(d <= k)
    {
        cout << d*a << endl;
        return ;
    }

    long long ans = k * a;
    d -= k;
    long long p = d / k;

    if(t + k * a < k * b) ans += (t + k * a) * p;
    else ans += k * b * p;

    d %= k;

    if(t + d * a < d * b) ans += (t + d * a);
    else ans += d * b;

    cout << ans << endl;
    //system("pause");
    return ;
}

这题只说一句：快速幂。

>技不如人，甘拜下风。

>相当精彩的比赛。

#include <iostream>
#include <string>
#include <map>
using namespace std;
int f[][], fa[];
long long SUM[][], MIN[][], w[];
long long Asum[], Amin[];
int Ap[];
int main()
{
    long long n, k;
    cin >> n >> k;
    for (int i = ; i < n; i++)
        cin >> fa[i];
    for (int i = ; i < n; i++)
        cin >> w[i];
    for (int i = ; i < n; i++)
    {
        f[i][] = fa[i];
        SUM[i][] = w[i];
        MIN[i][] = w[i];
    }
    for (int i = ; i <= ; i++)
    {
        for (int j = ; j < n; j++)
        {
            int x = f[j][i - ];
            f[j][i] = f[x][i - ];
            SUM[j][i] = SUM[j][i - ] + SUM[x][i - ];
            MIN[j][i] = MIN[j][i - ];
            if (MIN[j][i - ] > MIN[x][i - ]) MIN[j][i] = MIN[x][i - ];
        }
    }
    memset(Asum, , sizeof(Asum));
    for (int i = ; i < n; i++)
    {
        Amin[i] = 1e15;
        Ap[i] = i;
    }
    for (int i = ; i <= ; i++)
    {
        if ((k & ((long long) << i)) == ) continue;
        for (int j = ; j < n; j++)
        {
            int x = Ap[j];
            Asum[j] += SUM[x][i];
            if (Amin[j] > MIN[x][i]) Amin[j] = MIN[x][i];
            Ap[j] = f[x][i];
        }
    }
    for (int i = ; i < n; i++)
        cout << Asum[i] << " " << Amin[i] << endl;
    //system("pause");
    return ;
}