Educational Codeforces Round 39 F Largest Beautiful Number

传送门

暴力DP dp[i][j][k] 从j 到 k 发f(x) 的大小，

#include <bits/stdc++.h>
using namespace std;
typedef long long ll ;
typedef double    dl ;
#define INF 0x7fconst int  inf = 987654321;
const int sz = 1e6 + 5;
const int mod = 1e9 + 7;
const int sqrtn = 300;#define f(i,l,r) for(int i=l;i<=r;++i)
#define g(i,l,r) for(int i=l;i>=r;--i)
#define CLR(arr,val) memset(arr,val,sizeof(arr))
#define sz(a)   strlen(a)
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);#define lowbit(x) x&(-x)
#define X first
#define Y second #define pb push_back
#define pii pair<int,int>int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
int n,x;
string s;
ll dp [105][105][105];
ll powe[105];void work()
{cin>>n>>x>>s;powe[1]=2;powe[0]=2;f(i,2,104)powe[i]=powe[i-1]*powe[i-2]%mod;f(i,0, n-1) dp[s[i]-'0'][i][i]=1;f(i,2,x)f(j,0,n-1)f(k,j,n-1){if(k==n-1)dp[i][j][k]=(dp[i][j][k]+dp[i-1][j][k]*powe[i-2]%mod)%mod;elsedp[i][j][k]=(dp[i][j][k]+dp[i-1][j][k] )%mod;if (j == 0)dp[i][j][k] = (dp[i][j][k] + dp[i - 2][j][k]*powe[i - 1] % mod) % mod;elsedp[i][j][k] = (dp[i][j][k] + dp[i - 2][j][k]) % mod;f(l,j,k-1)dp[i][j][k] = (dp[i][j][k] +dp[i-1][j][l]*dp[i-2][l+1][k]%mod)%mod ;}cout << dp[x][0][n - 1] << endl;}int main()
{FAST_IO ;freopen("in","r",stdin);freopen("out","w",stdout);cout<<"hello"<<endl;work();return 0;
}

转载于:https://www.cnblogs.com/corx/p/8523876.html

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