预处理的广搜

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

49 8112 96 107 9

 

Sample Output

9No solution1122111112

 

题意：给出s1,s2。s1为位和，s2为位的平方和。求满足该s1和s2的最小数

思路：预处理广搜。从0 0 开始搜索、每个dp[i][j]只用进队一次。

#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <string>#include <string>using namespace std;#define maxn 108bool dp[902][8102];int pre[902][8102];int ans[102];struct Item{int step,ss1,ss2;Item(){}Item(int s,int s1,int s2){step = s;ss1 = s1;ss2 = s2;}};queue <Item> q;void init(){memset(dp,0,sizeof(dp));dp[0][0] = 0;memset(pre,-1,sizeof(pre));q.push(Item(0,0,0));while(!q.empty()){Item item = q.front();q.pop();if(item.step >= 100)break;for(int i = 1;i <= 9;i++){int nows1 = item.ss1 + i;int nows2 = item.ss2 + i*i;if(dp[nows1][nows2])continue;if(nows1 > 900 || nows2 > 8100) break;dp[nows1][nows2] = 1;pre[nows1][nows2] = i;q.push(Item(item.step+1,nows1,nows2));}}}int main(){//freopen("in.txt","r",stdin);init();int t;scanf("%d",&t);while(t--){int s1,s2;scanf("%d%d",&s1,&s2);if(s1 > 900 || s2 > 8100){printf("No solution\n");continue;}if(pre[s1][s2] == -1)printf("No solution\n");else{int pos = 0;while(pre[s1][s2] != -1){ans[pos++] = pre[s1][s2];int as = s1,bs = s2;s1 -= pre[as][bs];s2 -= pre[as][bs]*pre[as][bs];}for(int i = pos-1;i >= 0;i--)printf("%d",ans[i]);printf("\n");}}return 0;}