zoukankan      html  css  js  c++  java
  • SPOJ

    链接:

    https://vjudge.net/problem/SPOJ-BALNUM

    题意:

    Balanced numbers have been used by mathematicians for centuries. A positive integer is considered a balanced number if:

    1.  Every even digit appears an odd number of times in its decimal representation
      
    2.  Every odd digit appears an even number of times in its decimal representation
      

    For example, 77, 211, 6222 and 112334445555677 are balanced numbers while 351, 21, and 662 are not.

    Given an interval [A, B], your task is to find the amount of balanced numbers in [A, B] where both A and B are included.

    思路:

    三进制记录每个值用的奇数次还是偶数次。
    直接DP即可。

    代码:

    // #include<bits/stdc++.h>
    #include<iostream>
    #include<cstdio>
    #include<vector>
    #include<string.h>
    #include<set>
    #include<queue>
    #include<algorithm>
    #include<math.h>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    const int MOD = 1e9+7;
    const int MAXN = 1e6+10;
    
    ULL a, b;
    ULL F[21][60000];
    int dig[21];
    ULL m[11];
    
    int Upd(int x, int p)
    {
        int sum = 0;
        for (int i = 0;i < 10;i++)
        {
            int tmp = x%3;
            x /= 3;
            if (i == p)
                sum += (tmp == 1 ? 2 : 1) * m[i];
            else
                sum += tmp * m[i];
        }
        return sum;
    }
    
    bool Check(int x)
    {
        int p = 0;
        while(x)
        {
            if (x%3 == 2 && p%2 == 0)
                return false;
            if (x%3 == 1 && p%2 == 1)
                return false;
            x /= 3;
            p++;
        }
        return true;
    }
    
    ULL Dfs(int pos, int sta, bool zer, bool lim)
    {
        if (pos == -1)
            return Check(sta);
        if (!lim && F[pos][sta] != -1)
            return F[pos][sta];
        int up = lim ? dig[pos] : 9;
        ULL ans = 0;
        for (int i = 0;i <= up;i++)
        {
            ans += Dfs(pos-1, (zer && i == 0) ? 0 : Upd(sta, i), zer && i == 0, lim && i == up);
        }
        if (!lim)
            F[pos][sta] = ans;
        return ans;
    }
    
    ULL Solve(ULL x)
    {
        int p = 0;
        while(x)
        {
            dig[p++] = x%10;
            x /= 10;
        }
        return Dfs(p-1, 0, 1, 1);
    }
    
    int main()
    {
        // freopen("test.in", "r", stdin);
        m[0] = 1;
        for (int i = 1;i < 11;i++)
            m[i] = m[i-1]*3;
        memset(F, -1, sizeof(F));
        int t;
        scanf("%d", &t);
        while(t--)
        {
            scanf("%llu %llu", &a, &b);
            printf("%llu
    ", Solve(b)-Solve(a-1));
        }
    
        return 0;
    }
    
  • 相关阅读:
    php的session和cookie
    CRUD
    hibernate关系映射
    hibernate hql
    String和StringBuffer的区别
    策略模式Strategy
    项目结构
    final关键字
    项目中的建议
    struts学习记录
  • 原文地址:https://www.cnblogs.com/YDDDD/p/12000230.html
Copyright © 2011-2022 走看看