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  • Problem I. Increasing or Decreasing MIPT-2016 Pre-Finals Workshop, Taiwan NTU Contest, Sunday, March 27, 2016

    题面:

    Problem I. Increasing or Decreasing
    Input file: standard input
    Output file: standard output
    Time limit: 2 seconds
    Memory limit: 512 mebibytes
    We all like monotonic things, and solved many problems about that like Longest Increasing Subsequence
    (LIS). Here is another one which is easier than LIS (in my opinion).
    We say an integer is a momo number if its decimal representation is monotonic. For example, 123, 321,
    777 and 5566 are momo numbers; But 514, 50216 and 120908 are not.
    Please answer m queries. The i-th query is a interval [li; ri], and please calculate the number of momo
    numbers in it.
    Input
    The first line contains an integer m.
    Each of the following m lines contains two integers li; ri.
    1 m 105
    1 li ri 1018
    Output
    For each query, please output the number of momo numbers in that range.
    Example

    standard input standard output
    2
    1 100
    100 200
    100
    48


    思路:

      数位dp

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 #define MP make_pair
     4 #define PB push_back
     5 typedef long long LL;
     6 typedef pair<int,int> PII;
     7 const double eps=1e-8;
     8 const double pi=acos(-1.0);
     9 const int K=1e6+7;
    10 const int mod=1e9+7;
    11 
    12 LL dp[30][15][3];
    13 int num[30];
    14 LL dfs(int pos,int s,int pre,int led,int lmt)
    15 {
    16     if(pos<1) return !led;
    17     if(!lmt&&~dp[pos][pre][s]) return dp[pos][pre][s];
    18     int mx=lmt?num[pos]:9;
    19     LL ans=0;
    20     for(int i=0;i<=mx;i++)
    21     if(led)
    22         ans+=dfs(pos-1,0,i,!i&&led,lmt&&i==mx);
    23     else
    24     {
    25         if(s==0&&i>pre)
    26             ans+=dfs(pos-1,1,i,!i&&led,lmt&&i==mx);
    27         else if(s==0&&i<pre)
    28             ans+=dfs(pos-1,2,i,!i&&led,lmt&&i==mx);
    29         else if(s==0&&i==pre)
    30             ans+=dfs(pos-1,0,i,!i&&led,lmt&&i==mx);
    31         else if(s==1&&i>=pre)
    32             ans+=dfs(pos-1,1,i,!i&&led,lmt&&i==mx);
    33         else if(s==2&&i<=pre)
    34             ans+=dfs(pos-1,2,i,!i&&led,lmt&&i==mx);
    35     }
    36     //printf("lm:%d dp[%d][%d][%d]:%I64d
    ",lmt,pos,pre,s,ans);
    37     if(!lmt) dp[pos][pre][s]=ans;
    38     return ans;
    39 }
    40 LL go(LL x)
    41 {
    42     if(x<1) return 0;
    43     int cnt=0;
    44     LL ans=0;
    45     while(x)
    46         num[++cnt]=x%10,x/=10;
    47     for(int i=0;i<num[cnt];i++)
    48         ans+=dfs(cnt-1,0,i,!i,0);
    49     return ans+dfs(cnt-1,0,num[cnt],0,1);
    50 }
    51 
    52 int main(void)
    53 {
    54     LL m,l,r;
    55     cin>>m;
    56     memset(dp,-1,sizeof dp);
    57     while(m--)
    58     {
    59         scanf("%lld%lld",&l,&r);
    60         //printf("%I64d
    ",go(r));
    61         printf("%lld
    ",go(r)-go(l-1));
    62     }
    63     return 0;
    64 }
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  • 原文地址:https://www.cnblogs.com/weeping/p/7420966.html
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