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Digit Counting

2019-11-08 02:28:11
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Think: 這道題就是要求 求出1~~N 個數(shù)之間0 ~9出現(xiàn)的次數(shù)。 所以只需將每個數(shù)遍歷,將各位數(shù)字取出,并使其對應(yīng)的一維數(shù)組 +1 最后輸出即可。

Description Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 toN(1 < N < 10000) . After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, withN = 13 , the sequence is: 12345678910111213 In this sequence, 0 appears once, 1 appears 6 times, 2 appears 2 times, 3 appears 3 times, and each digit from 4 to 9 appears once. After playing for a while, Trung gets bored again. He now wants to write a PRogram to do this for him. Your task is to help him with writing this program. Input The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets. For each test case, there is one single line containing the number N . Output For each test case, write sequentially in one line the number of digit 0, 1,…9 separated by a space. Sample Input 2 3 13 Sample Output 0 1 1 1 0 0 0 0 0 0 1 6 2 2 1 1 1 1 1 1

題目大意 1. 輸入數(shù)據(jù)T代表下面有T組數(shù)據(jù) 2. 輸入數(shù)字K 表示求1~~~K 3. 要求輸出1~~K 各位數(shù)字的出現(xiàn)次數(shù)

#include<bits/stdc++.h>using namespace std;int main(){ int T; int k, i, t; int a[1050]; cin >>T; while(T --) { memset(a, 0, sizeof(a)); cin >> k; for (i = 1;i <= k;i ++) { int num = i; while(num) { t = num % 10; a[t] ++; num = num / 10; } } for (i = 0;i <= 9;i ++) { if (i == 9) cout << a[i] << endl; else cout << a[i] <<" "; } } return 0;}
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