gym101343 2017 JUST Programming Contest 2.0

A.On The Way to Lucky Plaza  (数论)
题意:m个店 每个店可以买一个小球的概率为p
       求恰好在第m个店买到k个小球的概率

题解:求在前m-1个店买k-1个球再*p就好了 最开始没太懂输出什么意思
       其实就是p*q的逆元的意思 因为概率是三位小数于是对他*1000*1000的逆元处理

　　 还要加个eps 因为0.005浮点数由于不确定性可能存的0.0050001或者0.00499999

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

ll ny[100005];

ll pow_mod(ll x, ll y)
{
    ll res = 1;
    while(y)
    {
        if(y & 1) res = res * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return res % mod;
}

int main()
{
    ll n, m, k;
    scanf("%lld%lld%lld", &m, &n, &k);
    for(ll i = 1; i <= 100000; i++) ny[i] = pow_mod(i, mod - 2LL);

    double pp;
    scanf("%lf", &pp);
    ll p = 1000 * pp + 1e-8;

    ll ans = 1;
    for(int i = 1; i <= k - 1; i++) ans = ans * ny[i] % mod;
    for(ll i = 1; i <= k - 1; i++) ans = ans * (n - i) % mod;
    for(int i = 1; i <= k; i++) ans = ans * p % mod;
    for(int i = 1; i <= n - k; i++) ans = ans * (1000 - p) % mod;
    for(int i = 1; i <= n; i++) ans = ans * ny[1000] % mod;
    printf("%lld
", ans);
    return 0;
}

B.So You Think You Can Count？(水DP）

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1000000007;

char s[10005];
int vis[10];
ll dp[10005];

int main()
{
    int n;
    scanf("%d", &n);
    scanf("%s", s + 1);
    dp[0] = 1;

    for(int i = 1; i <= n; i++)
    {
        memset(vis, 0, sizeof(vis));

        for(int j = i; j >= 1; j--)
        {
            if(vis[s[j] - '0']) break;
            else
            {
                vis[s[j] - '0'] = 1;
                dp[i] = (dp[i] + dp[j - 1]) % mod;
            }
        }
    }
    printf("%lld
", dp[n]);
    return 0;
}

C.MRT Map (最短路) 不会写最短路 学姐写的

D.Husam's Bug (水题)

E.Abdalrahman Ali Bugs (水题)

F.Certifications (水题)

G.In the Chairman's office (水题)

H.Give Me This Pizza

题意:给一个数组 求每个数右边第一个比他大的数是谁

题解:ai才不到50 从后往前暴力即可 存每个数最后出现的位置

#include <bits/stdc++.h>
using namespace std;

int vis[55];
int q[100005];
int ans[100005];

int main()
{
    int n;
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &q[i]);

    for(int i = 1; i <= 50; i++) vis[i] = 1000005;
    for(int i = 1; i <= n; i++) ans[i] = 1000005;

    for(int i = n; i >= 1; i--)
    {
        vis[q[i]] = i;
        int f = -1;
        for(int j = q[i] + 1; j <= 50; j++)
        {
            if(vis[j] < ans[i])
            {
                ans[i] = vis[j];
                f = j;
            }
        }
        ans[i] = f;
    }
    for(int i = 1; i <= n; i++)
    {
        if(i != n) printf("%d ", ans[i]);
        else printf("%d
", ans[i]);
    }
    return 0;
}

I.Husam and the Broken Present 1 (水题)

J.Husam and the Broken Present 2 (状压DP)  题解戳

K.Counting Time (模拟)

题意:给一个九宫格的初始状态 填完这个九宫格使得每个数字x能跳到x+1的方案有多少种

　　 能跳到的区域为八连通 且x只能跳到x+1

题解:模拟

#include <bits/stdc++.h>
using namespace std;

int nx[10];
int ny[10];
int vis[10];
char tu[5][5];
int dx[10];
int dy[10];
int q[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};

int ans;
bool check()
{
    dx[q[1]] = dx[q[2]] = dx[q[3]] = 1;
    dx[q[4]] = dx[q[5]] = dx[q[6]] = 2;
    dx[q[7]] = dx[q[8]] = dx[q[9]] = 3;
    dy[q[1]] = dy[q[4]] = dy[q[7]] = 1;
    dy[q[2]] = dy[q[5]] = dy[q[8]] = 2;
    dy[q[3]] = dy[q[6]] = dy[q[9]] = 3;

    for(int i = 1; i <= 9; i++)
        if(vis[i])
        {
            if(dx[i] == nx[i] && dy[i] == ny[i]) continue;
            else return false;
        }

    for(int i = 2; i <= 9; i++)
    {
        if(abs(dx[i] - dx[i - 1]) <= 1 && abs(dy[i] - dy[i - 1]) <= 1) continue;
        else return false;
    }
    return true;
}

int main()
{
    ans = 0;
    for(int i = 1; i <= 3; i++) scanf("%s", tu[i] + 1);
    for(int i = 1; i <= 3; i++)
        for(int j = 1; j <= 3; j++)
        {
            if(tu[i][j]!= '0')
            {
                int c = tu[i][j] - '0';
                vis[c] = 1;
                nx[c] = i;
                ny[c] = j;
            }
        }

    ans += check();
    while(next_permutation(q + 1, q + 10)) ans += check();
    printf("%d
", ans);
    return 0;
}