Codeforces 899D Shovel Sale (数学)

最新推荐文章于 2020-08-22 09:22:33 发布

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最新推荐文章于 2020-08-22 09:22:33 发布

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本文解析了CodeForces D.ShovelSale问题，介绍了一种算法来计算从n个铲子中选择两把铲子的方式，使得这两把铲子的价格总和以尽可能多的9结尾的方法数量。

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There are n shovels in Polycarp's shop. The i-th shovel costs i burles, that is, the first shovel costs 1 burle, the second shovel costs 2burles, the third shovel costs 3 burles, and so on. Polycarps wants to sell shovels in pairs.

Visitors are more likely to buy a pair of shovels if their total cost ends with several 9s. Because of this, Polycarp wants to choose a pair of shovels to sell in such a way that the sum of their costs ends with maximum possible number of nines. For example, if he chooses shovels with costs 12345 and 37454, their total cost is 49799, it ends with two nines.

You are to compute the number of pairs of shovels such that their total cost ends with maximum possible number of nines. Two pairs are considered different if there is a shovel presented in one pair, but not in the other.

Input

The first line contains a single integer n (2 ≤ n ≤ 109) — the number of shovels in Polycarp's shop.

Output

Print the number of pairs of shovels such that their total cost ends with maximum possible number of nines.

Note that it is possible that the largest number of 9s at the end is 0, then you should count all such ways.

It is guaranteed that for every n ≤ 109 the answer doesn't exceed 2·109.

   Examples
 
     input
   
7

     output
   
3

     input
   
14

     output
   
9

     input
   
50

     output
   
1

Note

In the first example the maximum possible number of nines at the end is one. Polycarp cah choose the following pairs of shovels for that purpose:

2 and 7;
3 and 6;
4 and 5.

In the second example the maximum number of nines at the end of total cost of two shovels is one. The following pairs of shovels suit Polycarp:

1 and 8;
2 and 7;
3 and 6;
4 and 5;
5 and 14;
6 and 13;
7 and 12;
8 and 11;
9 and 10.

In the third example it is necessary to choose shovels 49 and 50, because the sum of their cost is 99, that means that the total number of nines is equal to two, which is maximum possible for n = 50.

题目链接：http://codeforces.com/problemset/problem/899/D

题意：输入一个n，求从中任选两个数，使得其尾部9的个数最多的数对有多少个。（也可能存在0个9）

思路：分析数据可以发现，<50,尾部1个9；<500，尾部2个9；<5000，尾部3个9......依次类推。可以发现，只需要判断当前的数与5*10^x做比较即可判断尾部最多有多少个9。

设当前输入n的时候一对数尾部最多有x个9，设r为尾部有x个9的各个可能取值，

r > 2*n-1 (即r > n + (n-1)) break;

r < n+1 ans += r/2;

r < 2*n ans += (2*n-r+1)/2;

代码：

#include <bits/stdc++.h>
#define LL long long
using namespace std;

LL n,k,cnt,ans;

int main(){
    k = 5;
    scanf("%lld",&n);
    if(n < 5){
        printf("%d\n",n*(n-1)/2);
    }
    else{
        while(k * 10 <= n) k *= 10, cnt ++;
        LL temp = 1, num = 0;
        for(int i = 0; i <= cnt; i ++){
            num = num * 10 + 9;
            temp *= 10;
        }
        for(int i = 0; i <= 8; i ++){
            LL r = i * temp + num;
            if(r > 2*n-1) break;
            else if(r < n + 1) ans += r/2;
            else if(r < 2 * n) ans += (2*n-r+1)/2;
        }
        printf("%lld\n",ans);
    }
    return 0;
}