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  • Bomb(要49)--数位dp

    Bomb

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 25866    Accepted Submission(s): 9810


    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.
     
    #include<iostream>
    #include<string.h>
    #define ll long long
    using namespace std;
    ll shu[20], dp[20][2];
    ll dfs(ll len, bool if4, bool shangxian)
    {
        if (len == 0)
            return 1;
        if (!shangxian&&dp[len][if4])
            return dp[len][if4];
        ll mx, cnt = 0;//cnt记录的是区间内不含49的个数
        mx = (shangxian ? shu[len] : 9);
        for (ll i = 0; i <= mx; i++)
        {
            if (if4&&i == 9)//如果shu[len]==4&&上一个状态是9
                continue;
            cnt = cnt + dfs(len - 1, i == 4, shangxian&&i == mx);
        }
        return shangxian ? cnt : dp[len][if4] = cnt;
    }
    ll solve(ll n)
    {
        memset(shu, 0,sizeof(shu));
        ll k = 0;
        while (n)//将n的每一位拆解出来逆序存在shu[i]中。eg:109,shu[0]=9,shu[1]=0,shu[2]=1;
        {
            shu[++k] = n % 10;//注意这里是++k
            n = n / 10;
        }
        return dfs(k, false, true);
    }
    int main()
    {
        ll t;
        scanf("%lld", &t);
        while (t--)
        {
            ll n;
            scanf("%lld", &n);//这里计算的区间是[0,n],题目要计算的是[1,n];
            printf("%lld
    ", n-(solve(n)-1));
            //如果是计算区间[a,b];printf(solve(b)-solve(a-1));
    
        }
        return 0;
    
    }
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  • 原文地址:https://www.cnblogs.com/-citywall123/p/10745755.html
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