首页 > 代码库 > hud 3555 Bomb 数位dp
hud 3555 Bomb 数位dp
Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 19122 Accepted Submission(s): 7068
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?
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.Author
fatboy_cw@WHU
Source
2010 ACM-ICPC Multi-University Training Contest(12)——Host by WHU
数位dp的比较好的题目把。。
#include<stdio.h> #include<string.h> #include<algorithm> #include<iostream> #include<cmath> #define ls (u<<1) #define rs (u<<1|1) #define maxn 30 #define ll long long #define INF 1e18+7 using namespace std; #define max(a,b) (a)>(b)?(a):(b) #define min(a,b) (a)<(b)?(a):(b) int digit[maxn]; ll dp[maxn][3]; ll dfs(int pos,int flag,int limit){ if(pos == -1){ return flag == 2;//flag等于2的时候代表这个数满足条件 } if(!limit && dp[pos][flag]!=-1){//达到极限且此时dp有值 return dp[pos][flag]; } ll sum = 0; int e = limit?digit[pos]:9; for(int i=0;i<=e;i++){ int have = flag; if(flag == 1 && i == 9){ have = 2; } if(flag == 0 && i == 4){ have = 1;//此时为将要完成的状态 } if(flag == 1 && i!=4 && i!=9){ have = 0; } sum += dfs(pos-1,have,limit&&i==e); } if(!limit){//达到极限的情况 dp[pos][flag] = sum; } return sum; } ll solve(ll n){ int pos = 0; while(n){ digit[pos++] = n%10; n /= 10; } return dfs(pos-1,0,1); } int main(){ int T; scanf("%d",&T); while(T--){ ll n; scanf("%lld",&n); memset(dp,-1,sizeof(dp)); printf("%lld\n",solve(n)); } return 0; }
hud 3555 Bomb 数位dp
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。