zoukankan      html  css  js  c++  java
  • Bomb

    Bomb
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    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
    分析:数位dp;
    代码:
    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cmath>
    #include <algorithm>
    #include <climits>
    #include <cstring>
    #include <string>
    #include <set>
    #include <map>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <list>
    #define rep(i,m,n) for(i=m;i<=n;i++)
    #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
    #define mod 1000000007
    #define inf 0x3f3f3f3f
    #define vi vector<int>
    #define pb push_back
    #define mp make_pair
    #define fi first
    #define se second
    #define ll long long
    #define pi acos(-1.0)
    #define pii pair<int,int>
    #define Lson L, mid, ls[rt]
    #define Rson mid+1, R, rs[rt]
    const int maxn=1e5+10;
    using namespace std;
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
    inline ll read()
    {
        ll x=0;int 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;
    }
    int m,k,t,d[20];
    ll n,dp[20][3];
    ll dfs(int now,int ca,int ok)
    {
        if(now<0)return ca==2;
        if(ok&&dp[now][ca]!=-1)return dp[now][ca];
        int len=ok?9:d[now];
        ll ans=0;
        for(int i=0;i<=len;i++)
        {
            if(ca==0)
            {
                if(i==4)ans+=dfs(now-1,1,ok||i<len);
                else ans+=dfs(now-1,0,ok||i<len);
            }
            else if(ca==1)
            {
                if(i==9)ans+=dfs(now-1,2,ok||i<len);
                else if(i==4)ans+=dfs(now-1,1,ok||i<len);
                else ans+=dfs(now-1,0,ok||i<len);
            }
            else if(ca==2)
            {
                ans+=dfs(now-1,2,ok||i<len);
            }
        }
        if(ok)dp[now][ca]=ans;
        return ans;
    }
    ll solve(ll p)
    {
        int num=0;
        memset(dp,-1,sizeof(dp));
        while(p)
        {
            d[num++]=p%10;
            p/=10;
        }
        return dfs(num-1,0,0);
    }
    int main()
    {
        int i,j;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%lld",&n);
            printf("%lld
    ",solve(n));
        }
        //system("Pause");
        return 0;
    }
  • 相关阅读:
    常用SQL Server数据修复命令DBCC一览(转载)
    多线程编程的注意事项
    [转载]Openlayers中使用TileCache加载预切割图片作为基础地图图层
    一个关于工具条可以移动和在四周停留的测试
    SQL Server同名表的判断
    SQL Server 中调整自增字段的当前初始值
    IIS and ASP.NET: The Application Pool
    Server.MapPath(path)
    自动化selenium 鼠标键盘操作& 按键用法
    多层窗口定位&多层框架定位
  • 原文地址:https://www.cnblogs.com/dyzll/p/5932605.html
Copyright © 2011-2022 走看看