PAT 解题报告 1049. Counting Ones (30)

1049. Counting Ones (30)

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
给定一个正整数 N(<=230)，要求计算所有小于 N 的正整数的各个位置上，1 出现的次数之和。

分析
比较有思维难度的一题，核心在于找规律。10ms 的时间限制表明了不能用常规的循环遍历来解决。需要从简单的 case 找规律，逐步扩大到常规的情况。

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
//规律0~9有1个1；0~99有20个1；0~999有300个1...
using namespace std;
#define UP 10
int a[UP] = {0};
void mka(){
for (int i=1; i<UP; i++) {
double tmp = pow(10, i-1);
a[i] = tmp * i;
}
}
int getD(int n) {//得出几位数，123是3位数
int cnt = 0;
while (n!=0) {
n/=10;
cnt++;
}
return cnt;
}

int main()
{
mka();
int N;
scanf("%d", &N);
int sum = 0;
while (N !=0 ) {//每次循环处理最高位
if (N<10 && N>=1) {
sum += 1;
break;
}
int d = getD(N);//d位数
double tmp = pow(10, d-1); int t = tmp;
int x = N / t;//最高位的数字
if (x ==1) {
sum += N - x*t + 1;
}
else if (x>1) {
sum += t;
}
sum += x*a[d-1];
N -= x*t;//去掉最高位，处理后面的数
}
printf("%d", sum);

return 0;
}