Apocalypse Someday poj 3208

题目

问你在所有包含666的数中，第n大的是多少。(1 ≤ n ≤ 50,000,000) 。开头几个是666, 1666, 2666, 3666, 4666, 5666…

分析

套路
先dp预处理，再一位位的填数。

code

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<vector>#define N 50
#define LL long longusing namespace std;long long f[N][4];void dp()
{f[0][0]=1;for (int i=1;i<=20;i++){f[i][0]=9*f[i-1][0]+9*f[i-1][1]+9*f[i-1][2];f[i][1]=f[i-1][0];f[i][2]=f[i-1][1];f[i][3]=f[i-1][2]+10*f[i-1][3];}
}int T,ans;
int main()
{freopen("data.in","r",stdin);freopen("data.out","w",stdout);scanf("%d",&T);dp();while (T--){int x; scanf("%d",&x);int m;for (m=1;m<=20;m++)if (f[m][3]>=x) break;long long num=0;int k=0;for (int i=m;i>=1;i--){int j;for (j=0;j<=9;j++){if (j==6||k==3){long long cnt=0;if (j==6) cnt=f[i-1][3];for (int l=max(3-k-(j==6),0);l<3;l++)cnt+=f[i-1][l];if (cnt+num<x) num+=cnt;else break;}if (j!=6) {if (f[i-1][3]+num<x) num+=f[i-1][3];else break;}}printf("%d",j);if (k<3){if (j==6) k++;else k=0;}}printf("\n");}
}

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Apocalypse Someday poj 3208

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分析

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