PAT A1049题解


1049 Counting Ones (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:给定一个数,列举1-n,统计其中的1的个数


题目思路:一开始没有注意到时间限制,所以当然第一反应是进行n^2复杂度的两重循环列举,但是明显这样是不符合题意,定然超时的。所以一开始写的是这样的。


//部分超时代码
#includecstdio
#includeiostream
using namespace std;
int change(int n,int a[]){
	int count=0;
	while(n!=0){
		a[count++]=n%10;
		n=n/10;
	}
}
int main(){
	int n,m,sum=0;
	cinn;
	for(int i=1;i=n;i++){
		int a[10]={0};
		change(i,a);
		for(int j=0;j10;j++){
			if(a[j]==1){
				sum++;
			}
		}
	}
	coutsum;
	return 0;
}

在考虑到时间限制当中,还是得要找到数学上的规律才能符合题意。


//AC 代码
#include iostream
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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