zoukankan      html  css  js  c++  java
  • VF(动态规划)

    VF

    时间限制:1000 ms  |  内存限制:65535 KB
    难度:2
    描述
    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 = 109) because Vasya himself won’t cope with the task. Can you solve the problem?
    输入
    There are multiple test cases.
    Integer S (1 ≤ S ≤ 81).
    输出
    The milliard VF value in the point S.
    样例输入
    1
    样例输出
    10
    描述:题意较难理解,就是小于1000000000的个位数字合为n的个数,动态规划;
    #include<stdio.h>
    int main(){
        int vf[10][90]={0},i,j,k,sum,s;
        
        for(int i=1;i<=9;++i)vf[1][i]=1;
        for(int i=1;i<=9;++i)for(int j=1;j<=9*i;++j)for(int k=0;k<=j&&k<=9;++k)vf[i][j]+=vf[i-1][j-k];
        while(~scanf("%d",&s)){sum=0;
        if(s==1)printf("10
    ");
        else{
            for(int i=1;i<=9;++i)sum+=vf[i][s];
            printf("%d
    ",sum);}
        }
        return 0;
    }
    
  • 相关阅读:
    JVM 源码分析
    GGGGCCCC
    正则化(Regularization)、过拟合(Overfitting)
    名校课程
    数据库垂直拆分 水平拆分
    运维角度浅谈MySQL数据库优化
    表的垂直拆分和水平拆分
    Eclipse去掉对JS文件的Validation
    Linux定时任务工具crontab详解及系统时间同步
    高性能分布式哈希表FastDHT
  • 原文地址:https://www.cnblogs.com/handsomecui/p/4529281.html
Copyright © 2011-2022 走看看