hdu3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 8554    Accepted Submission(s): 2981

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.

 

 

怎么到处都是英文题……

题意是求1到n中含有“49”子串的数字的个数。多组询问。

这一看就是数位dp啊……（话说如果不只一个限制的话岂不是ac自动机上dp）

预处理:f[i][j]表示i位的数（允许前导0），头一位是j的含"49"的方案数

显然f[i][j]=Σf[i-1][k],0<=k<9.

当第i位是4而第i-1位是9的时候，已经满足条件，因此后面取什么都无所谓了。直接加上第i-2位以后的所有数的个数，即10^(i-2)。

由容斥原理，f[i][j]再减去原来多加的f[i-1][9]。

当处理的时候，先把当前询问的x一位一位提出来。

1、首先考虑当前不存在a[now+1]==4且a[now]==9的情况：注意到此时方案是不会相互影响的，也就是取1000~1999的方案数和取51000~51999的方案数是一样的。因为在头上加上5之后并不会造成额外的"49"产生。

2、如果a[now+1]==4且a[now]==9，显然跟上面一样，接下来取什么都是无所谓的。直接加上x剩下来的后now-1位数rest

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define pi 3.1415926535897932384626433832795028841971
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void write(LL a)
{
    if (a<0){printf("-");a=-a;}
    if (a>=10)write(a/10);
    putchar(a%10+'0');
}
inline void writeln(LL a){write(a);printf("
");}
int n;
LL f[30][10];
LL sec[20];
inline LL wrk(LL x)
{
    LL tot=0,dat=x;
    int dig[20]={0},len=0;
    while (x)
    {
        dig[++len]=x%10;
        x/=10;
    }
    while (len)
    {
        dat-=dig[len]*sec[len-1];
        for (int i=0;i<dig[len];i++)
        tot+=f[len][i];
        if (dig[len+1]==4&&dig[len]==9)
        {
            tot+=dat+1;
            break;
        }
        len--;
    }
    return tot;
}
int main()
{
    n=read();
    sec[0]=1;for(int i=1;i<=17;i++)sec[i]=sec[i-1]*10;
    for (int i=2;i<=20;i++)
    for (int j=0;j<10;j++)
    {
        for (int k=0;k<10;k++)f[i][j]+=f[i-1][k];
        if (j==4)f[i][j]+=sec[i-2]-f[i-1][9];
    }
    for(int i=1;i<=n;i++)
    {
        LL x=read();
        writeln(wrk(x));
    }
    return 0;
}