HDU 5867 Water problem 【】

Water problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1051    Accepted Submission(s): 441

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

3123

 

Sample Output

3611

 

Author

BUPT

 

Source

2016 Multi-University Training Contest 10

 

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))const int M=1e6+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;int i,j,k,n,m;int a[100]={0,3,3,5,4,4,3,5,5,4,3,6,6,8,8,7,7,9,8,8,6};int main(){a[30]=6;a[40]=5;a[50]=5;a[60]=5;a[70]=7;a[80]=6;a[90]=6;    int t;    scanf("%d",&t);    while(t--){        scanf("%d",&m);        n=0;        ll ans=0;        for(int i=1;i<=m;i++){            n=i;          if(n<=20)ans+=a[n];          else if(n<100){            int sum=0;            sum+=a[(n/10)*10];            sum+=a[n%10];            ans+=sum;           }           else if(n%100==0&&n<1000){                ans+=7;                ans+=a[n/100];           }           else {              if(n==1000){ans+=11;}              else {               int sum=0;               sum+=a[n/100];               sum+=10;               int kk=n-100*(n/100);               if(kk<=20)sum+=a[kk];               else {                 int sum=0;                 sum+=a[(kk/10)*10];                  sum+=a[kk%10];                 ans+=sum;              }               ans+=sum;            }        }        }        printf("%lld\n",ans);    }    return 0;}