zoukankan      html  css  js  c++  java
  • 第一轮 M

    Milliard Vasya's Function
    Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
    Submit Status
    
    Description
    Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most interesting facts such as Pythagor’s theorem are already proved? Correct! He is to think out something his own, original. So he thought out the Theory of Vasya’s Functions. Vasya’s Functions (VF) are rather simple: the value of the Nth VF in the point S is an amount of integers from 1 to N that have the sum of digits S. You seem to be great programmers, so Vasya gave you a task to find the milliard VF value (i.e. the VF with N = 10 9) because Vasya himself won’t cope with the task. Can you solve the problem?
    
    Input
    Integer S (1 ≤ S ≤ 81).
    
    Output
    The milliard VF value in the point S.
    
    Sample Input
    input	output
    
    1
    
    	
    
    10
    
    /*************************************************************************
    	> File Name: m.cpp
    	> Author: yuan
    	> Mail:
    	> Created Time: 2014年11月10日 星期一 22时24分14秒
     ************************************************************************/
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cstdlib>
    #include<stdlib.h>
    #include<algorithm>
    #include<cmath>
    #define MAX 1000000000
    using namespace std;
    char str[15];
    //int ans[82];
    int main()
    {
       /* int t=MAX;
        printf("%d
    ",t);
        memset(ans,0,sizeof(ans));
        for(int i=1;i<=MAX;i++)
        {
           memset(str,0,sizeof(str));
            sprintf(str,"%d",i);
            int l=strlen(str);
            int sum=0;
            for(int j=0;j<l;j++)
            {
                sum+=str[j]-'0';
            }
            ans[sum]++;
        }
        for(int i=1;i<=81;i++)
        {
            printf(",%d",ans[i]);
        }*/
        int ans[82]={0,10,45,165,495,1287,3003,6435,12870,24310,43749,75501,125565,202005,315315,478731,708444,1023660,1446445,2001285,2714319,3612231,4720815,6063255,7658190,9517662,11645073,14033305,16663185,19502505,22505751,25614639,28759500,31861500,34835625,37594305,40051495,42126975,43750575,44865975,45433800,45433800,44865975,43750575,42126975,40051495,37594305,34835625,31861500,28759500,25614639,22505751,19502505,16663185,14033305,11645073,9517662,7658190,6063255,4720815,3612231,2714319,2001285,1446445,1023660,708444,478731,315315,202005,125565,75501,43749,24310,12870,6435,3003,1287,495,165,45,9,1};
        int ll;
        while(~scanf("%d",&ll)){
            printf("%d
    ",ans[ll]);
        }
        return 0;
    }
    


  • 相关阅读:
    哈希表
    跳表
    哈夫曼之谜
    选择树、判定树和查找树

    将gbk字符串转换成utf-8,存储到注册表中后,再次从注册表读取转换成gbk,有问题!!!
    函数内部还是不要使用 strtok()
    没想到: System.out.println(n1 == f1 ? n1 : f1);
    在不同DPI屏幕环境下,让图标显示的尺寸保持不变,使用 LoadImage() 加载图标
    在多线程中显示模态窗口,出现异常现象
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254395.html
Copyright © 2011-2022 走看看