题目链接：
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Problem Description
The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?
Input
The first line of input consists of an integer T (1
<= T <= 10000), indicating the number of test cases. For each test case,
there will be an integer N (1 <= N <= 2^63-1) as the
description.
The input terminates by end of file marker.
The input terminates by end of file marker.
Output
For each test case, output an integer indicating the
final points of the power.
Sample Input
3
1
50
500
Sample Output
0
1
15
Hint
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499",
so the answer is 15.题意：求出[1,n]中多少个数包含49；
AC代码：
#include<bits/stdc++.h> using namespace std; typedef long long ll; int dig[20]; ll dp[20][3]; /* dp[pos][0]:长度为pos的数中，不包含49的，前一位不为4的有多少个； dp[pos][1]:长度为pos的数中，不包含49的，前一位为4的有多少个； dp[pos][2]:长度为pos的数中，包含49的有多少个； */ ll dfs(int pos,int pre,bool have49,bool limit) { if(pos==0) return have49; if(!limit)//首先满足没有上界限制 { if(have49 && ~dp[pos][2]) return dp[pos][2]; if(!have49 && pre==4 && ~dp[pos][1]) return dp[pos][1]; if(!have49 && pre!=4 && ~dp[pos][0]) return dp[pos][0]; } int up=limit?dig[pos]:9; ll ans=0; for(int i=0;i<=up;i++) { if(pre==4 && i==9) ans+=dfs(pos-1,i,1,limit && i==up); else ans+=dfs(pos-1,i,have49,limit && i==up); } if(!limit) { if(have49) dp[pos][2]=ans; else { if(pre==4) dp[pos][1]=ans; else dp[pos][0]=ans; } } return ans; } ll solve(ll x) { int pos=0; while(x) { dig[++pos]=x%10; x/=10; } return dfs(pos,0,0,1); } int main() { int T; ll N; scanf("%d",&T); while(T--) { scanf("%I64d",&N); memset(dp,-1,sizeof(dp)); printf("%I64d\n",solve(N)); } }