Codeforces_822

A.小的那个数的阶乘。

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

int a,b;

int main()
{
    ios::sync_with_stdio();
    cin >> a >> b;
    int t  = min(a,b);
    int ans = ;
    for(int i = ;i <= t;i++)    ans *= i;
    cout << ans << endl;
    return ;
}

---

B.暴力t串的起始位置，比较两个串。

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

int n,m;
string s1,s2;
vector<int> v;

int main()
{
    ios::sync_with_stdio();
    cin >> n >> m >> s1 >> s2;
    s1 = ' '+s1;
    s2 = ' '+s2;
    int ans = 1e9;
    for(int i = ;i <= m-n;i++)
    {
        int cnt = ;
        for(int j = ;j <= n;j++)
        {
            if(s1[j] != s2[i+j])    cnt++;
        }
        if(cnt < ans)
        {
            ans = cnt;
            v.clear();
            for(int j = ;j <= n;j++)
            {
                if(s1[j] != s2[i+j])    v.push_back(j);
            }
        }
    }
    cout << ans << endl;
    for(int i = ;i < v.size();i++) cout << v[i] << " ";
    cout << endl;
    return ;
}

---

C.先按l排序，遍历每一段，维护一个优先队列，每次把当前之前的段从优先队列中拿出来，更新对应长度的最小花费，然后把该段放进优先队列。

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

int n,x;
map<int,int> mp;
struct xx
{
    int l,r,c;
    friend bool operator <(xx a,xx b)
    {
        return a.l < b.l;
    }
}a[];
struct yy
{
    int r,len,c;
    yy(int a,int b,int cc):r(a),len(b),c(cc){};
    friend bool operator <(yy a,yy b)
    {
        return a.r > b.r;
    }
};
priority_queue<yy> q;

int main()
{
    ios::sync_with_stdio();
    cin >> n >> x;
    for(int i = ;i <= n;i++)   cin >> a[i].l >> a[i].r >> a[i].c;
    sort(a+,a++n);
    int ans = 2e9+;
    for(int i = ;i <= n;i++)
    {
        while(!q.empty() && q.top().r < a[i].l)
        {
            if(!mp.count(q.top().len))  mp[q.top().len] = q.top().c;
            else    mp[q.top().len] = min(mp[q.top().len],q.top().c);
            q.pop();
        }
        int len = a[i].r-a[i].l+;
        if(mp.count(x-len)) ans = min(ans,a[i].c+mp[x-len]);
        q.push(yy(a[i].r,len,a[i].c));
    }
    if(ans == 2e9+)    cout << - << endl;
    else    cout << ans << endl;
    return ;
}

---

D.暴力dp刚好卡过。

#include<bits/stdc++.h>
#define MOD 1000000007
using namespace std;

int t,l,r;
long long dp[],f[];

int main()
{
    ios::sync_with_stdio();
    cin >> t >> l >> r;
    for(int i = ;i <= r;i++)
    {
        f[i] = (long long)i*(i-)/;
        dp[i] = f[i];
    }
    for(int i = ;i <= r;i++)
    {
        for(int j = i,k = ;j <= r;j += i,k++)  dp[j] = min(dp[j],dp[i]*k+f[k]);
    }
    long long tt = ,ans = ;;
    for(int i = l;i <= r;i++)
    {
        ans = (ans+dp[i]%MOD*tt)%MOD;
        tt = tt*t%MOD;
    }
    cout << ans << endl;
    return ;
}

---

E.dp，类似01背包，开两个数组记录两个串中的位置。

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

string s,t;
int n,m,x,a[],b[];

int main()
{
    ios::sync_with_stdio();
    cin >> n >> s >> m >> t >> x;
    s = ' '+s;
    t = ' '+t;
    for(int i = ;i <= x;i++)   a[i] = ;
    for(int i = ;i <= n;i++)
    {
        for(int j = x;j >= ;j--)
        {
            int now = a[j];
            if(i < b[j]) continue;
            int k = ;
            while(k+now <= m)
            {
                if(s[i+k] != t[now+k])  break;
                k++;
            }
            if(now+k > m)
            {
                cout << "YES" << endl;
                return ;
            }
            if(a[j+] < now+k)
            {
                a[j+] = now+k;
                b[j+] = i+k;
            }
        }
    }
    cout << "NO" << endl;
    return ;
}

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