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  • PAT 1049 Counting Ones[dp][难]

    1049 Counting Ones (30)(30 分)

    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 (<=2^30^).

    Output Specification:

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

    Sample Input:

    12

    Sample Output:

    5

     题目大意:给出一个数字N,你要找出所有1-N中包含的1的个数。N<=2^30。

     //N还挺大的。猛一看很简答,直接遍历就行,但是数据量太大。如果使用动态规划,化解成子问题,怎么做呢?

     代码来自:https://www.liuchuo.net/archives/2305

    我还不太懂,明天再看一遍这数学问题。

    #include <iostream>
    #include<stdio.h>
    using namespace std;
    int main() {
        int n, left = 0, right = 0, a = 1, now = 1, ans = 0;
        scanf("%d", &n);
        while(n / a) {
            left = n / (a * 10), now = n / a % 10, right = n % a;
            if(now == 0) ans += left * a;
            else if(now == 1) ans += left * a + right + 1;
            else ans += (left + 1) * a;
            a = a * 10;
        }
        printf("%d", ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/BlueBlueSea/p/9475250.html
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