洛谷2019 3月月赛 T4

T3做不来。。 直接滚去T4 orz

乍一看 T4是个DP

题干

复杂度？？（N^4） 咋优化。。。

还带一只捆绑 捆绑啥的最烦人了

最后20pts 直接废了 T了 很烦

不过拿到80pts已经很开心了惹

#pragma GCC optimize(1)
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
inline LL read () {
    LL res =  ;
    int f () ;
    char ch = getchar ();
    while (!isdigit(ch)) {
        if (ch == '-') f = - ;
        ch = getchar();
    }
    while (isdigit(ch)) res = (res << ) + (res << ) + (ch ^ ),ch = getchar();
    return res * f ;
}
const int N=<<,Mod=1e9+;
int n;
int f[N*N][N][N];
namespace slove {
    void Init() {
        n=read();
        f[][][] = ;
        for (register int i = ; i<n*(n-); i++)
            for (register int j=; j<=((i+>n)?n:i+); j++)
                for (register int k=j; k<=((i+>n)?n:i+); k++)
                    if (f[i][j][k]) {
                        if (i<=k+j- and k<n) f[i+][j][k+]=(f[i+][j][k+]+f[i][j][k])%Mod;
                        else if(((k*(k-)+j*(j-))>>)>=i+) f[i+][j][k]=(f[i+][j][k]+f[i][j][k])%Mod;
                        for (register int l=; l<=k-j; l++) if (((k*(k-)+(j+l)*(j+l-))>>)>=i+) f[i+][j+l][k]=(f[i+][j+l][k]+f[i][j][k])%Mod;
                    }
    }
    void slove() {
        Init();
        for (register int i=; i<=n*n-n; i++) {
            LL ans=;
            for (register int j =; j<=((i+>n)?n:i+); j++)
                for (register int k=j; k<=((i+>n)?n:i+); k++) ans=(ans+f[i][j][k])%Mod;
            printf("%d%c",ans," \n"[i==n*n-n]);
        }
    }
}
signed main() {
    return slove::slove(),;
}

关于100pts ↓

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

template <class T>
inline T Min(const T &a, const T &b) {return a < b ? a : b;}

const int N = , M = N * N, ZZQ = 1e9 + ;

int n, p_limit[N], f[][N][N], sf[][N][N], g[][N], sg[][N], ans[N * N];

int main()
{
    std::cin >> n;
    for (int i = ; i <= n; i++)
        p_limit[i] = (n - i + ) * (n - ) + (i - ) * (i - ) / ;
    f[][n][] = ans[] = ;
    for (int i = ; i <= n; i++) sf[][i][] = ;
    for (int i = ; i <= Min(n * (n - ), n << ); i++)
    {
        int op = i & ;
        for (int j = ; j <= n; j++)
            for (int k = ; k <= n; k++)
                f[op][j][k] = ;
        for (int j = ; j <= n; j++) if (i <= p_limit[j])
            for (int k = ; k <= n; k++) if (i + j >= n + k - )
                f[op][j][k] = (f[op ^ ][j][k] + sf[op ^ ][j + ][k - ]) % ZZQ;
        for (int j = n; j >= ; j--)
            for (int k = ; k <= n; k++)
            {
                sf[op][j][k] = (sf[op][j + ][k] + f[op][j][k]) % ZZQ;
                ans[i] = (ans[i] + f[op][j][k]) % ZZQ;
            }
    }
    for (int j = ; j <= n; j++) for (int k = ; k <= n; k++)
        g[][j] = (g[][j] + f[][j][k]) % ZZQ;
    for (int j = n; j >= ; j--) sg[][j] = (sg[][j + ] + g[][j]) % ZZQ;
    for (int i = (n << ) + ; i <= n * (n - ); i++)
    {
        int op = i & ;
        for (int j = ; j <= n; j++) g[op][j] = ;
        for (int j = ; j <= n; j++) if (i <= p_limit[j])
            g[op][j] = sg[op ^ ][j];
        for (int j = n; j >= ; j--)
        {
            sg[op][j] = (sg[op][j + ] + g[op][j]) % ZZQ;
            ans[i] = (ans[i] + g[op][j]) % ZZQ;
        }
    }
    for (int i = ; i <= n * (n - ); i++) printf("%d ", ans[i]);
    puts("");
    return ;
}