杭电 3555 Bomb

Bomb


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6609    Accepted Submission(s): 2303




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.
 


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.



第一个数位动规!

!!!

看代码理解的。差点儿相同懂那么点点点了。


AC代码例如以下:

传统模板!

///递推   46MS 328K

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;


int main()
{
    __int64 dp[20][3];//0表示该位数中不含49的情况数。1表示不含49但头为9的情况数。2表示含49的情况数
    int i,j;
    memset(dp,0,sizeof dp);
    dp[0][0]=1;
    for(i=1;i<=20;i++)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
        dp[i][1]=dp[i-1][0];
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
        //cout<<dp[i][0]<<" "<<dp[i][1]<<" "<<dp[i][2]<<endl;
    }
    int t;
    cin>>t;
    __int64 num[30];
    __int64 n;
    while(t--)
    {
        memset(num,0,sizeof num);
        cin>>n;
        n++;//这一步非常重要
        int tt=1;
        while(n)
        {
            num[tt++]=n%10;
            n/=10;
        }
        __int64 ans=0;
        int flag=0;int last=0;
        for(i=tt-1;i>=1;i--)
        {
            ans+=num[i]*dp[i-1][2];
            if(flag==1)
                ans+=dp[i-1][0]*num[i];
            if(flag==0&&num[i]>4)
                ans+=dp[i-1][1];
            if(num[i+1]==4&&num[i]==9)
                flag=1;
        }
        cout<<ans<<endl;
    }
    return 0;
}



///数位之记忆化搜索   78MS 276K


#include<cstdio>
#include<iostream>
#include<cstring>
#define mod 1000000009
#define ll __int64
#define M 100005
using namespace std;

ll dp[30][3];
ll num[30];

ll dfs(int pos,int statu,int limit)
{
    int i;
    if(!pos) return statu==2;
    if(!limit&&dp[pos][statu]!=0) return dp[pos][statu];
    int end=limit?

num[pos]:9; ll sum=0; for(i=0;i<=end;i++) { int flag=statu; if(flag==0&&i==4) flag=1; if(flag==1&&i==9) flag=2; if(flag==1&&i!=4&&i!=9) flag=0; sum+=dfs(pos-1,flag,limit&&i==end); } return limit?

sum:(dp[pos][statu]=sum); } ll _49(ll n) { int pos=1; memset(dp,0,sizeof dp); while (n>0) { num[pos++]=n%10; n/=10; } //cout<<pos<<"~~~~~~~~~~~~~"<<endl; return dfs(pos-1,0,1); } int main() { int i,j; int t; scanf("%d",&t); while(t--) { ll n; scanf("%I64d",&n); printf("%I64d\n",_49(n)); } return 0; }



posted @ 2017-05-23 11:21  yfceshi  阅读(148)  评论(0编辑  收藏  举报