#308 (div.2) B. Vanya and Books
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1.题目描述:点击打开链接
2.解题思路:本题要求统计数位的个数,简单的试验一下发现有如下规律:一个n位数的个数有9*(10^n)个,因此所有n位数的数位是n*9*(10^n)个,因此可以利用两个循环变量base,k来计算,其中base表示n位数的总个数,k表示每一个n位数的数位有k位,循环条件是n-base>0,这样即可完成统计。
3.代码:
#define _CRT_SECURE_NO_WARNINGS#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;typedef long long ll;typedef unsigned long long ull;#define me(s) memset(s,0,sizeof(s))#define For(i,n) for(int i=0;i<(n);i++)#define pb push_back#define sz size#define clr clear#define F(a,b) for(int i=a;b;i++)int main(){ int n; while(~scanf("%d",&n)) { ll cnt=0; ll base=9,k=1; while(n-base>0) { n-=base; cnt+=base*k; base*=10; k++; } cnt+=n*k; printf("%I64d\n",cnt); } return 0;}
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