【刷题】uva10954 Add All【3】

【刷题】uva10954 Add All【3】

• 一、题目
  • 1.题目描述
• 二、解题报告
  • 1.思路分析
  • 2.代码详解
  • 3.注意事项

一、题目

1.题目描述

题目：Fahim likes to solve mathematical problems. But sometimes it becomes challenging for him to solve all the mathematical problems. So sometimes he gets angry to solve mathematical puzzles. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000). After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, with N = 13, the sequence is. 12345678910111213 The sequence is interesting, right! In this sequence, zero(0) appears once, one(1) appears 6 times, two(2) appears 2 times, three(3) appears 3 times, and each digit from four(4) to nine(9) appears once. After playing for a while, Fahim 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. 输入：Look at the input file carefully. It 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. 输出：Now for each individual test case, write sequentially in one line the number of digit 0, 1, . . . 9 separated by a space. 示例 :

二、解题报告

1.思路分析

• 通过不断取模的方式直到取到模为0为止，得到数字的每一位数t5%10
• 创建数组0-9的全零数组，通过判断数组索引t2==t5%10来累计计数

2.代码详解

C++👇

#include<iostream>
int main(){
	int k=0;
	scanf("%d",&k);
	for (int t1=0;t1<k;t1++){
		int N=0;
		int result[10]={0};
		scanf("%d",&N);
		for (int t2=0;t2<10;t2++){
			for(int t3=1;t3<N+1;t3++){
				int t5=0;
				t5=t3;
				while(t5!=0){
					if(t2==t5%10){
						result[t2]++;
					}
					t5=t5/10;
				}
			
			}
		}
		for(int t4=0;t4<10;t4++){
			printf("%d",result[t4]);
			if(t4!=9){
				printf(" ");
			} else{
				printf("\n");
			}
		}
	
	}
} 

3.注意事项

1.数组脏数据

这样👇

2.输出的格式一定是后面直接回车，不能多空格