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:
    报表中的Excel操作之Aspose.Cells(Excel模板)
    .NET开源组件
    JSON 和 JSONP
    servlet 中getLastModified()
    spring mvc源码-》MultipartReques类-》主要是对文件上传进行的处理,在上传文件时,编码格式为enctype="multipart/form-data"格式,以二进制形式提交数据,提交方式为post方式。
    spring mvc dispatcherservlet处理request流程
    log显示error时的堆栈信息理解和分析
    web项目log日志查看分析->流程理解
    war包结构
    Spring Boot干货系列:(三)启动原理解析
  • 原文地址:https://www.cnblogs.com/codeyuan/p/4254395.html
Copyright © 2011-2022 走看看