HDU 5867-Water problem（单词字母个数）

超、凢脫俗 2022-09-25 11:18 220阅读 0赞

# Water problem #

**Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)  
Total Submission(s): 254 Accepted Submission(s): 168**

Problem Description

If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3+3+5+4+4=19 letters used in total.If all the numbers from 1 to n (up to one thousand) inclusive were written out in words, how many letters would be used?  
  
Do not count spaces or hyphens. For example, 342 (three hundred and forty-two) contains 23 letters and 115 (one hundred and fifteen) contains 20 letters. The use of "and" when writing out numbers is in compliance with British usage.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases.  
  
For each test case: There is one positive integer not greater one thousand.

Output

For each case, print the number of letters would be used.

Sample Input

3
    1
    2
    3

Sample Output

3
    6
    11

Author

BUPT

Source

[2016 Multi-University Training Contest 10][]

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wange2014

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    #define maxn 1010
    #define INF 0xfffffff
    using namespace std;
    
    
    int main()
    {
        int i,j;
        /**0~100**/
        int a[maxn]= {0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8,6};
        a[30]=6,a[40]=5,a[50]=5,a[60]=5,a[70]=7,a[80]=6,a[90]=6,a[100]=10,a[1000]=11;
        for(i=200; i<=900; i+=100)
            a[i]=7+a[i/100];
    
        for(j=20; j<=90; j+=10)
            for(i=1+j; i<=9+j; ++i)
                a[i]=a[j]+a[i-j];
        /* for(i=0; i<=100; ++i)
             cout<<"i="<<i<<" "<<a[i]<<endl;
         cout<<endl;*/
        /**100~200**/
        /*for(i=1+100; i<=200; ++i)
            a[i]=a[100]+3+a[i-100];
         for(i=101; i<=250; ++i)
             cout<<"i="<<i<<" "<<a[i]<<endl;
         cout<<endl;*/
        /**100~1000**/
        for(j=100; j<=900; j+=100)
            for(i=1+j; i<=99+j; ++i)
                a[i]=a[j]+3+a[i-j];
        /*for(i=1; i<=1000; ++i)
            cout<<"i="<<i<<" "<<a[i]<<endl;
        cout<<endl;*/
        int t;
        long long ans=0;
        scanf("%d",&t);
        while(t--)
        {
            int n;
            scanf("%d",&n);
            ans+=a[n];
            printf("%I64d\n",ans);
        }
        return 0;
    }
    /*
    3
    1
    2
    3
    */

[2016 Multi-University Training Contest 10]: http://acm.split.hdu.edu.cn/search.php?field=problem&key=2016+Multi-University+Training+Contest+10&source=1&searchmode=source