Gym 101147补题

D - Popcorn（排列组合）

就是求\(C^x_y\)
代码1：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
#define ll long long 

void init(){
      cin.tie();
      cout.tie();
      std::ios::sync_with_stdio(false);
}
int main()
{  
     // freopen("popcorn.in","r",stdin);
    init();
    int n; scanf("%d",&n);
       for(int i=0;i<n;i++){
       ll ans[25];
        ll x,y;
        scanf("%lld %lld",&x,&y);
        ans[0]=1;
        for(int i=1;i<=x;i++){
            ans[i]=ans[i-1]*(x-i+1)/i;
            cout<<ans[i]<<endl;
        }
        printf("%lld\n",ans[y]);
       // cout<<ans[y]<<endl;
    }
    return 0;
}

代码2：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
#define ll long long 
void init(){
      cin.tie();
      cout.tie();
      std::ios::sync_with_stdio(false);
}
int main()
{  init();
    freopen("popcorn.in","r",stdin);
    ll n;cin>>n;
    for(ll i=0;i<n;i++){
        ll x,y;cin>>x>>y;
        ll ans=1;
        for(ll i=x;i>=x-y+1;i--){
            ans*=i;
        }
       for(ll i=y;i>=1;i--){
            ans/=i;
        }
        cout<<ans<<endl;
    }
}

H - Commandos Gym - 101147H

简单dp

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
#define ll long long
const int N = 11;

int a[N][N][N];
int dp[N][N];
int main()
{

 //    freopen("commandos.in","r",stdin);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        memset(dp, 0, sizeof dp);
        memset(a, 0, sizeof a);
        for (int i = 0; i < n; i++)
        {
            int x, y, z, w;
            cin >> x >> y >> z >> w;
            a[x][y][z] += w;
        }
        for (int i = 10; i >= 1; i--)
        {
            for (int j = 1; j <= 10; j++)
            {
                for (int o = 1; o <= 10; o++)
                {
                    dp[j][o]=max(dp[j][o]+ a[i][j][o],
                max(dp[j - 1][o] + a[i][j][o],dp[j][o - 1] + a[i][j][o]));
                //     dp[j][o]=max(dp[j][o], );

                    //       if(j==5&&i==10){cout<<a[i][j][o]<<" "<<dp[j][o]<<endl;}
                }
            }

        }

        cout << dp[10][10] << endl;
    }

    return 0;
}

E - Jumping Gym - 101147E 变形最短路

单边，从最后向前遍历，因为这样可以一遍得到从n->（1~n-1）的最短路

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5 + 10;
#define pii pair<int, int>
const int INF = 0x3f3f3f3f;
vector<pii> g[N];
int dis[N];
int vis[N];
int n;
void dj()
{
    priority_queue<pii, vector<pii>, greater<pii>> q;
    memset(dis, 0x3f, sizeof dis);
 //   memset(vis, 0, sizeof vis);
    q.push({0, n});
    dis[n] = 0;
    while (q.size())
    {
        pii k = q.top();
        q.pop();
     //   if (vis[k.second])
      //      continue;
     //   vis[k.second] = 1;
        for (pii t : g[k.second])
        {

            int dist = k.first + t.first;
            if (dis[t.second] > dist)
            {
                dis[t.second] = dist;
                q.push({dist, t.second});
            }
        }
    }

    for (int i = 1; i <= n; i++)
    {
        if (dis[i] >= INF)
        {
            printf("-1\n");
        }
        else
            printf("%d\n", dis[i]);
    }
}
int main()
{

    // freopen("jumping.in","r",stdin);

    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        for (int i = 0; i < N; i++)
        {
            g[i].clear();
        }
        for (int i = 1; i <= n; i++)
        {
            int t;
            scanf("%d", &t);
            if (i - t >= 1)
            {
               // g[i].push_back({1, i - t});
                g[i - t].push_back({1, i});
            }
            if (i + t <= n)
            {
             //   g[i].push_back({1, i + t});
                g[i + t].push_back({1, i});
            }
        }
        dj();
    }
    return 0;
}