Digit Generator UVa1583

最新推荐文章于 2024-02-20 13:43:04 发布

Devinxtw

最新推荐文章于 2024-02-20 13:43:04 发布

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本文链接：https://blog.youkuaiyun.com/qq_40688707/article/details/80141301

本文介绍了一种算法，用于寻找特定整数的最小生成器，即找到一个正整数N，使得N及其各位数字之和等于给定的目标整数。通过预计算方式加速查找过程，并提供了一个C++实现示例。

For a positive integer N, the digit-sum of N is defined as the sum of N itself and its digits. When Mis the digitsum of N, we call N a generator of M.For example, the digit-sum of 245 is 256 (= 245 + 2 + 4 + 5). Therefore, 245 is a generator of256.Not surprisingly, some numbers do not have any generators and some numbers have more than onegenerator. For example, the generators of 216 are 198 and 207.You are to write a program to find the smallest generator of the given integer.InputYour program is to read from standard input. The input consists of T test cases. The number of testcases T is given in the first line of the input. Each test case takes one line containing an integer N,1 ≤ N ≤ 100, 000.OutputYour program is to write to standard output. Print exactly one line for each test case. The line is tocontain a generator of N for each test case. If N has multiple generators, print the smallest. If N doesnot have any generators, print ‘0’.Sample Input32161212005Sample Output19801979

#include <iostream>
#include<stdio.h>
#include <string.h> 
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
using namespace std;
int ans[100005];
//int table(int n)
//{
//	memset(ans , 0,sizeof(ans));
//	for(int m = 1; m < 100005;m++)
//	{
//		int x = m,y = m;
//		while(x > 0)
//		{
//			y += x % 10;
//			x /= 10;
//		}
//		if(ans[y] == 0 || m < ans[y])
//		ans[y] = m;
//			
//	}return ans[n];
//}
int main(int argc, char** argv) {
	for(int m = 1; m < 100005;m++)
	{
		int x = m,y = m;
		while(x > 0)
		{
			y += x % 10;
			x /= 10;
		}
		if(ans[y] == 0 || m < ans[y])
		ans[y] = m;
			
	}
	memset(ans , 0,sizeof(ans));
	for(int m = 1; m < 100005;m++)
	{
		int x = m,y = m;
		while(x > 0)
		{
			y += x % 10;
			x /= 10;
		}
		if(ans[y] == 0 || m < ans[y])
		ans[y] = m;
    }
	int n,t;
	scanf("%d",&n);
	while(n--)
	{
		
		scanf("%d",&t);
		printf("%d\n",ans[t]/*table(n)*/);
	}
	
	return 0;
}

这个题使用封装函数会超时，