http://poj.org/problem?id=2000

#include<stdio.h>
const int N=;
int main()
{
int coin[N];
long long n,i = ,j,k;
j = ;
k = ;
while (i < )
{
if (k < j)
{
coin[i++] = j;
k++;
}
if (k >= j)
{
k = ;
j++;
} }
while(~scanf("%lld",&n)&&n)
{
long long scoin = ;
for (i = ; i <= n; i ++)
scoin += coin[i];
printf("%lld %lld\n",n,scoin);
}
return ;
}

Gold Coins的更多相关文章

  1. Gold Coins 分类: POJ 2015-06-10 15:04 16人阅读 评论(0) 收藏

    Gold Coins Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21767   Accepted: 13641 Desc ...

  2. OpenJudge/Poj 2000 Gold Coins

    1.链接地址: http://bailian.openjudge.cn/practice/2000 http://poj.org/problem?id=2000 2.题目: 总Time Limit: ...

  3. hdoj 2401 Baskets of Gold Coins

    Baskets of Gold Coins Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  4. HDOJ(HDU) 2401 Baskets of Gold Coins(数列、)

    Problem Description You are given N baskets of gold coins. The baskets are numbered from 1 to N. In ...

  5. Baskets of Gold Coins

    Baskets of Gold Coins Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Ot ...

  6. H - Gold Coins(2.4.1)

    H - Gold Coins(2.4.1) Crawling in process... Crawling failed Time Limit:1000MS     Memory Limit:3000 ...

  7. poj 2000 Gold Coins(水题)

    一.Description The king pays his loyal knight in gold coins. On the first day of his service, the kni ...

  8. poj 2000 Gold Coins

    题目链接:http://poj.org/problem?id=2000 题目大意:求N天得到多少个金币,第一天得到1个,第二.三天得到2个,第四.五.六天得到3个....以此类推,得到第N天的金币数. ...

  9. Baskets of Gold Coins_暴力

    Problem Description You are given N baskets of gold coins. The baskets are numbered from 1 to N. In ...

随机推荐

  1. 51nod 1118 机器人走方格【dp】

    M * N的方格,一个机器人从左上走到右下,只能向右或向下走.有多少种不同的走法?由于方法数量可能很大,只需要输出Mod 10^9 + 7的结果. 收起 输入 第1行,2个数M,N,中间用空格隔开.( ...

  2. [Algorithm] 11. Linked List Cycle

    Description Given a linked list, determine if it has a cycle in it. To represent a cycle in the give ...

  3. HDU - 5952 Counting Cliques(dfs搜索)

    题目: A clique is a complete graph, in which there is an edge between every pair of the vertices. Give ...

  4. db2层级查询

    CREATE VIEW v_orgtype99 asSELECT t1.SYS_ORG_TYPE_NAME top_name1, t2.SYS_ORG_TYPE_NAME top_name2, --若 ...

  5. 调用subprocess调用shell命令时屏蔽标准输出

    import os, subprocessp = subprocess.Popen(args, stdout = subprocess.PIPE,stderr = subprocess.STDOUT)

  6. 2018.5.7每天一题面试题----final, finally, finalize 的区别

    1.final修饰符(关键字).被final修饰的类,就意味着不能再派生出新的子类,不能作为父类而被子类继承. 因此一个类不能既被abstract声明,又被final声明.将变量或方法声明为final ...

  7. qemu-guest-agent简介

    经常使用vmWare的同学都知道有vmware-tools这个工具,这个安装在vm内部的工具,可以实现宿主机与虚拟机的通讯,大大增强了虚拟机的性能与功能, 如vmware现在的Unity mode下可 ...

  8. BNUOJ 14381 Wavio Sequence

    Wavio Sequence Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UVA. Origina ...

  9. 【POJ3159】Candies(差分约束系统)

    题意:有一些人, 给n个人派糖果,给出m组约束,每组约束包含A,B,c 三个数, 意思是A的糖果数比B少的个数不多于c,即B的糖果数 - A的糖果数<= c . 最后求n 比 1 最多多多少糖果 ...

  10. NOIP2015 提高组合集

    NOIP 2015 提高组 合集 D1 T1 神奇的幻方 题目让你干啥你就干啥,让你咋走你就咋走就完事儿了 #include <iostream> #include <cstdio& ...