Hdu3022 Sum of Digits

Sum of Digits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 810    Accepted Submission(s): 220

Problem Description

Petka thought of a positive integer n and reported to Chapayev the sum of its digits and the sum of its squared digits. Chapayev scratched his head and said: "Well, Petka, I won't find just your number, but I can find the smallest fitting number." Can you do the same?

Input

The first line contains the number of test cases t (no more than 10000). In each of the following t lines there are numbers s1 and s2 (1 ≤ s1, s2 ≤ 10000) separated by a space. They are the sum of digits and the sum of squared digits of the number n.

Output

For each test case, output in a separate line the smallest fitting number n, or "No solution" if there is no such number or if it contains more than 100 digits.

Sample Input

4
9 81
12 9
6 10
7 9

Sample Output

9
No solution
1122
111112

Source

2009 Multi-University Training Contest 6 - Host by WHU

题目大意：求一个数字，使得这个数字每个数位上的数字和为s1,平方和为s2,输出最小的满足这个要求的数字，如果不存在，则输出No solution

分析：好题！

　　　显然是一个dp.状态的每一维都很好确定，但它具体表示什么呢？ 这就比较头疼了.令f[i][j]表示和为i,平方和为j的数的最小位数. g[i][j]表示和为i,平方和为j,最小位数为f[i][j]的最小首位数. 如果能求得这两个数组，每次输出答案的时候先输出g[s1][s2],然后s1 -= g[s1][s2],s2 -= g[s1][s2],直到s1和s2中有一个等于0.

　　　怎么转移呢？f的转移非常简单，g的定义涉及到f,不好单独处理.  一个比较好的方法是把f和g放在一起处理. 每当f能转移的时候,就转移g.比如f[i][j]转移到f[i + k][j + k * k],那么和为i + k,j + k * k的最小位数在这个时候肯定是确定的，就是f[i + k][j + k * k],因为k是从小到大枚举的，所以g[i + k][j + k * k]也可以转移.g[j + k][j + k * k] = k. 如果f[i + k][j + k * k] == f[i][j] + 1, g的条件是满足了，但是最小首位数不一定是k,因为之前求出了f[i+k][j + k * k]是从其它的状态转移过去的，这个时候取个min.

　　　这道题的状态表示真的挺神奇的. 状态表示的东西必须要能够得到答案和转移,并且还要满足题目的要求（最小）. 考虑如何使得数最小，先是数位最少，再是首位最小.根据这两个最小就可以定义得到状态了.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

int T,s1,s2,f[][],g[][];

void solve()
{
    for (int i = ; i <= ; i++)
        f[i][i * i] = ,g[i][i * i] = i;
    for (int i = ; i <= ; i++)
        for (int j = ; j <= ; j++)
            if (f[i][j])
            {
                for (int k = ; k <= ; k++)
                {
                    if (!f[i + k][j + k * k] || f[i + k][j + k * k] > f[i][j] + )
                    {
                        f[i + k][j + k * k] = f[i][j] + ;
                        g[i + k][j + k * k] = k;
                    }
                    else if (f[i + k][j + k * k] == f[i][j] + )
                        g[i + k][j + k * k] = min(g[i + k][j + k * k],k);
                }
            }
}

int main()
{
    solve();
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d%d",&s1,&s2);
        if (s1 >  || s2 >  || !f[s1][s2] || f[s1][s2] > )
            printf("No solution\n");
        else
        {
            while (s1 && s2)
            {
                printf("%d",g[s1][s2]);
                int t = g[s1][s2];
                s1 -= t;
                s2 -= t * t;
            }
            printf("\n");
        }
    }

    return ;
}