HDU-3555 Bomb （数位DP）

Bomb

http://acm.hdu.edu.cn/showproblem.php?pid=3555

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

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.

题目大意：统计1~n中含有连续49的数字的个数？

数位DP入门题

设dp[i][0]表示长度为i，不含49的数字个数；dp[i][1]表示长度为i，不含49且最高位为9的数字个数；dp[i][2]表示长度为i，含有49的数字个数

不是很明白为什么第二维状态要这样取...

#include 

using namespace std;

int num[25],len;
long long n,ans;
long long dp[25][3];//dp[i][0]表示长度为i，不含49的数字个数；dp[i][1]表示长度为i，不含49且最高位为9的数字个数；dp[i][2]表示长度为i，含有49的数字个数

void Init() {
    dp[0][0]=1;
    dp[0][1]=dp[0][2]=0;
    for(int i=1;i<=20;++i) {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];//长度为i-1且不含49的，最高位加上0~9，共10种情况；但若最高位加上4，则要减去次高位为9的情况dp[i-1][1]
        dp[i][1]=dp[i-1][0];//长度i-1且不含49的，最高位加上9
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];//长度i-1且含49的，最高位加上0~9；长度i-1且最高位为9的，最高位加上4
    }
}

long long getCnt(long long x) {//得到区间[1,x]中满足题意的数字个数
    ++x;
    ans=len=0;
    while(x>0) {
        num[++len]=x%10;
        x/=10;
    }
    num[len+1]=-1;
    bool flag=false;//判断高位是否出现过49，true表示高位出现过
    for(int i=len;i>=1;--i) {//从最高位开始枚举
        ans+=dp[i-1][2]*num[i];//当前位取0~(num[i]-1)，后面i-1位数为含49的数

        if(flag) {//如果高位已经出现过49
            ans+=dp[i-1][0]*num[i];//当前位去0~(num[i]-1)，后面i-1位数为不含49的数【最开始已经加上了含49的数，所以不必再加】
        }
        if(!flag&&num[i]>4) {//如果高位未出现过49，且当前位为大于4
            ans+=dp[i-1][1];
        }
        if(num[i+1]==4&&num[i]==9) {//若该位与上一位能组成49
            flag=true;
        }
    }
    return ans;
}

int main() {
    Init();
    int T;
    scanf("%d",&T);
    while(T-->0) {
        scanf("%I64d",&n);
        printf("%I64d\n",getCnt(n));
    }
    return 0;
}