HDOJ3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 17591    Accepted Submission(s): 6473

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

3150500

 

Sample Output

0115

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.

法一（数位DP

#include <bits/stdc++.h>#define mst(a) memset(a,0,sizeof(a))#define f(i,a,b) for(int i=a;i<b;i++)using namespace std;long long dp[20][3];void init(){    mst(dp);    dp[0][2]=1;    f(i,1,20)    {        dp[i][0]=dp[i-1][0]*10+dp[i-1][1]; //含49        dp[i][1]=dp[i-1][2]; //不含49但最高位为9        dp[i][2]=dp[i-1][2]*10-dp[i-1][1]; //不含49且最高位不为9    }}long long fun(long long n){    int len=1;    int bit[25];    while(n)    {        bit[len++]=n%10;        n/=10;    }    bit[len]=0;    int flag=0;    long long sum=0;    for(int i=len;i>0;i--)    {        sum+=bit[i]*dp[i-1][0];        if(flag)            sum+=bit[i]*dp[i-1][2];        if(!flag&&bit[i]>4)            sum+=dp[i-1][1];        if(bit[i+1]==4&&bit[i]==9)            flag=1;    }    return sum;}int main(){    init();    int t;    long long n;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        printf("%I64d\n",fun(n+1));    }    return 0;}

法二（记忆化搜索）

#include <bits/stdc++.h>#define mst(a) memset(a,-1,sizeof(a))#define f(i,a,b) for(int i=a;i<=b;i++)using namespace std;int bit[20];long long dp[20][3];//0:没有49且最后一位不为4，1：没有49且最后一位为4，2：有49//flag=1;无限制，flag=0，有限制long long dfs(int pos,int st,int flag){    if(pos==0) return st==2;    if(flag&&dp[pos][st]!=-1) return dp[pos][st];    long long sum=0;    int x=flag?9:bit[pos];    f(i,0,x)    {        if((st==2)||(st==1&&i==9))            sum+=dfs(pos-1,2,flag||i<x);        else if(i==4) sum+=dfs(pos-1,1,flag||i<x);        else sum+=dfs(pos-1,0,flag||i<x);    }    if(flag) dp[pos][st]=sum;    return sum;}long long cal(long long x){    int len=0;    while(x)    {        bit[++len]=x%10;        x/=10;    }    return dfs(len,0,0);}int main(){    int t;    scanf("%d",&t);    mst(dp);    while(t--)    {        long long n;        scanf("%I64d",&n);        printf("%I64d\n",cal(n));    }    return 0;}