HDU 5867

题意

one:3个字母，two:3个字母，three:5个字母…..求到n<=1000，1—1000所有单词字母数量的和

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.

思路

大模拟，把特殊单词处理好就ok

AC代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int getonenum( int n ){ //19以内
    switch(n){
        case 1 : return 3;  case 2 : return 3;  case 3 : return 5;
        case 4 : return 4;  case 5 : return 4;  case 6 : return 3;
        case 7 : return 5;  case 8 : return 5;  case 9 : return 4;
        case 10: return 3;  case 11: return 6;  case 12: return 6;
        case 13: return 8;  case 14: return 8;  case 15: return 7;
        case 16: return 7;  case 17: return 9;  case 18: return 8;
        case 19: return 8;
    }
}

int zhengshi( int n ){  //整十的 20 -- 90
    switch(n){
        case 2 : return 6;  case 3 : return 6;  case 4 : return 5;
        case 5 : return 5;  case 6 : return 5;  case 7 : return 7;
        case 8 : return 6;  case 9 : return 6;
    }
}

int gettwonum( int n ){
    int a = n / 10, b = n % 10;
    if( n <= 19 ) return getonenum(n);
    else if( b == 0 ) return zhengshi(a);
    else return zhengshi(a)+getonenum(b);
}

int getnum(int n){
    //hunderd == 7
    if( n <= 19 ) return getonenum(n);
    else if( n < 100 ) return gettwonum(n);
    else if( n == 100 ) return 10;
    else if( n < 1000 ){
        int x = n;
        int c = x % 10; x /= 10;
        int b = x % 10; x /= 10;
        int a = x;
        //cout << a << ' ' << b << ' ' << c << endl;
        if( c == 0 && b == 0 ){
            return getonenum(a)+7;
        }
        else if( b*10+c <= 19 ){
            return getonenum(a)+7+3+getonenum(b*10+c);
        }
        else if( c == 0 ){
            return getonenum(a)+7+3+zhengshi(b);
        }
        else{
            return getonenum(a)+7+3+zhengshi(b)+getonenum(c);
        }
    }
    else if( n == 1000 ) return 11; //one thousand
}

int solve( int n ){
    int sum = 0;
    for( int i = 1; i <= n; i++ ){
        sum += getnum(i);
    }
    return sum;
}

int main()
{
    int T, n;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        printf("%d
",solve(n));
    }
    return 0;
}