PAT (Advanced Level) Practise 1049 Counting Ones (30)

1049. Counting Ones (30)

时间限制

100 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (<=2³⁰).

Output Specification:

For each test case, print the number of 1's in one line.

Sample Input:

12

Sample Output:

5

题意：给你一个n，问1~n中一共出现了几个1

解题思路：可以发现0~9出现了1个1，0~99出现了10+1*10个1，0~99出现了100+20*10个1......然后可以根据这个规律去打表，之后就可以根据这个表去进行模拟

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <cmath>#include <map>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;LL a[16],n,x[16];void init(){    a[0]=0,a[1]=1;    LL x=10;    for(int i=2;i<16;i++) a[i]=x+a[i-1]*10,x*=10;}int main(){    init();    while(~scanf("%lld",&n))    {        int cnt=0;        LL ans=0,m=n,k=1;        while(n) {x[cnt++]=n%10,n/=10,k*=10;}        k/=10;        for(int i=cnt-1;i>=0;i--)        {            ans+=a[i]*x[i];            if(x[i]==1) ans+=(m-k*x[i]+1);            if(x[i]>1) ans+=k;            m-=(k*x[i]);            k/=10;        }        printf("%lld\n",ans);    }    return 0;}