POJ题目2282 The Counting Problem（数学）

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概述

The Counting Problem

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 3559		Accepted: 1890

Description

Given two integers a and b, we write the numbers between a and b, inclusive, in a list. Your task is to calculate the number of occurrences of each digit. For example, if a = 1024 and b = 1032, the list will be
1024 1025 1026 1027 1028 1029 1030 1031 1032
there are ten 0's in the list, ten 1's, seven 2's, three 3's, and etc.

Input

The input consists of up to 500 lines. Each line contains two numbers a and b where 0 < a, b < 100000000. The input is terminated by a line `0 0', which is not considered as part of the input.

Output

For each pair of input, output a line containing ten numbers separated by single spaces. The first number is the number of occurrences of the digit 0, the second is the number of occurrences of the digit 1, etc.

Sample Input

Sample Output

1 2 1 1 1 1 1 1 1 1
85 185 185 185 190 96 96 96 95 93
40 40 40 93 136 82 40 40 40 40
115 666 215 215 214 205 205 154 105 106
16 113 19 20 114 20 20 19 19 16
107 105 100 101 101 197 200 200 200 200
413 1133 503 503 503 502 502 417 402 412
196 512 186 104 87 93 97 97 142 196
398 1375 398 398 405 499 499 495 488 471
294 1256 296 296 296 296 287 286 286 247

Source

Shanghai 2004

题目大意：输入两个数a，b，求a~b之间所有数0~9出现的次数

ac代码

#include<stdio.h>
#include<string.h>
int ansa[20],ansb[20];
void dfs(int n,int *ans,int t)
{
	int i,a,b;
	if(n<=0)
		return;
	a=n%10;
	b=n/10;
	for(i=1;i<=a;i++)
		ans[i]+=t;
	while(b)
	{
		ans[b%10]+=t*(a+1);
		b/=10;
	}
	for(i=0;i<=9;i++)
		ans[i]+=t*(n/10);
	t*=10;
	dfs((n/10)-1,ans,t);
}
int main()
{
	int a,b;
	while(scanf("%d%d",&a,&b)!=EOF,a||b)
	{
		int i;
		if(a>=b)
		{
			int t=a;
			a=b;
			b=t;
		}
		memset(ansa,0,sizeof(ansa));
		memset(ansb,0,sizeof(ansb));
		a--;
		if(b>a)
		{
			dfs(a,ansa,1);
			dfs(b,ansb,1);
		}
		for(i=0;i<=9;i++)
		{
			if(i)
				printf(" %d",ansb[i]-ansa[i]);
			else
				printf("%d",ansb[i]-ansa[i]);
		}
		printf("n");
	}
}