URAL 1658. Sum of Digits(簡單dp)

畫船聽雨發表於2014-05-20
Git

給你n，m。然後輸出100位以內一個數字，每一位的和為n，每一位的平方和為m。

以n，m為dp[i][j]。pre[i][j]記錄位置上的數字。

1658. Sum of Digits

Time limit: 2.0 second
Memory limit: 64 MB
Petka thought of a positive integer n and reported to Chapaev the sum of its digits and the sum of its squared digits. Chapaev 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 tlines 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 output 
4
9 81
12 9
6 10
7 9

9
No solution
1122
111112
 
#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-8
#define M 1000100
#define LL __int64
///#define LL long long
#define INF 0x3f3f3ff
#define PI 3.1415926535898
#define MOD 1000000009

const int maxn = 10100;

using namespace std;

int dp[910][8110];
int pre[910][8110];

void Pre()
{
    for(int i = 0; i <= 900; i++)
        for(int j = 0; j <= 8100; j++)
            dp[i][j] = 101;
    dp[0][0] = 0;
    for(int i = 1; i <= 900; i++)
    {
        for(int j = 1; j <= 8100; j++)
        {
            for(int k = 1; k<=9&&k<=i&&k*k<=j; k++)
            {
                if(i-k > j-k*k)
                    break;
                if(dp[i][j] > dp[i-k][j-k*k]+1)
                {
                    dp[i][j] = dp[i-k][j-k*k]+1;
                    pre[i][j] = k;
                }
            }
        }
    }
}

int main()
{
    Pre();
    int n, m;
    int T;
    cin >>T;
    while(T--)
    {
        scanf("%d %d",&n, &m);
        if(n>m ||  n > 900 || m > 8100 || dp[n][m] > 100 )
        {
            cout<<"No solution"<<endl;
            continue;
        }
        int t;
        while(n&&m)
        {
            t = pre[n][m];
            printf("%d",t);
            n -= t;
            m -= t*t;
        }
        cout<<endl;
    }
    return 0;
}


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