hdu3555——Bomb（数位DP）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 3212    Accepted Submission(s): 1142

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.
 
Author
fatboy_cw@WHU

Source
2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

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zhouzeyong

解析： 

         和hdu2089一样，就不多说了。。。

         很基础的数位DP

代码：

#include<cstdio>#include<algorithm>using namespace std;//dp[i][0]没有出现49//dp[i][1]没有出现49且最高位为9//dp[i][2]出现了49long long dp[21][3];void init(){    memset(dp,0,sizeof(dp));    dp[0][0]=1;    for(int i=1;i<20;i++)    {        dp[i][0]=(long long)dp[i-1][0]*10-dp[i-1][1];        dp[i][1]=dp[i-1][0];        dp[i][2]=dp[i-1][1]+(long long)dp[i-1][2]*10;    }}void read(){    freopen("hdu3555.in","r",stdin);    freopen("hdu3555.out","w",stdout);    int test;    scanf("%d",&test);    while(test--)    {        long long x;        scanf("%I64d",&x);        x++;    long long y=x;  int l=0;   int wei[30];    bool flag=0;    while(y)       {        wei[++l]=(y%10);       y/=10;    }    long long ret=0;    wei[l+1]=0;    for(int i=l;i>=1;i--)    {        ret+=(long long)dp[i-1][2]*wei[i];        if(flag)ret+=dp[i-1][0]*wei[i];        if(!flag&&wei[i]>4)ret+=dp[i-1][1];        if(wei[i+1]==4&&wei[i]==9)flag=true;    }    printf("%I64d\n",ret);    }}int main(){    init();    read();    return 0;}