您的位置：首页 > 其它

URAL 1658. Sum of Digits(简单dp)

2014-05-20 17:17 435 查看
给你n，m。然后输出100位以内一个数字，每一位的和为n，每一位的平方和为m。

以n，m为dp[i][j]。pre[i][j]记录位置上的数字。


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

inputoutput
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
[m] > 100 )
{
cout<<"No solution"<<endl;
continue;
}
int t;
while(n&&m)
{
t = pre
[m];
printf("%d",t);
n -= t;
m -= t*t;
}
cout<<endl;
}
return 0;
}
内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 点击这里给我发消息
标签： 
相关文章推荐
新的分享
章节导航