Digital Square
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1882 Accepted Submission(s): 741
Problem Description
Given an integer N,you should come up with the minimum nonnegative integer M.M meets the follow condition: M2%10x=N (x=0,1,2,3....)
Input
The first line has an integer T( T< = 1000), the number of test cases.
For each case, each line contains one integer N(0<= N <=109), indicating the given number.
For each case, each line contains one integer N(0<= N <=109), indicating the given number.
Output
For each case output the answer if it exists, otherwise print “None”.
Sample Input
3
3
21
25
Sample Output
None
11
5
Source
解析：设M = abc，即M = 100*a+10*b+c，则M2 = 10000*a2 + 1000*2*a*b + 100*(b2+2*a*c) + 10*2*b*c + c2。易知，M2的个位只与M的最后1位有关，M2的十位只与M的最后2位有关，M2的百位只与M的最后3位有关……我们可以从个位开始进行广搜，逐位满足N。如果有可行解，找出这一层当中最小的解并输出；否则输出"None"。
1 #include <cstdio> 2 #include <queue> 3 using namespace std; 4 5 struct node{ 6 long long value,place; 7 }; 8 9 void bfs(long long n) 10 { 11 node tmp; 12 tmp.value = 0; 13 tmp.place = 1; 14 queue<node> q; 15 q.push(tmp); 16 bool findans = false; 17 long long ans = 0xffffffff; 18 while(!q.empty()){ 19 tmp = q.front(); 20 q.pop(); 21 node now; 22 now.place = tmp.place*10; 23 for(int i = 0; i<10; ++i){ 24 now.value = tmp.value+i*tmp.place; 25 if(now.value*now.value%now.place == n%now.place){ 26 if(!findans) 27 q.push(now); 28 if(now.value*now.value%now.place == n && now.value<ans){ 29 findans = true; 30 ans = now.value; 31 } 32 } 33 } 34 } 35 if(ans == 0xffffffff) 36 printf("None "); 37 else 38 printf("%I64d ",ans); 39 } 40 41 int main() 42 { 43 int t; 44 scanf("%d",&t); 45 while(t--){ 46 long long n; 47 scanf("%I64d",&n); 48 bfs(n); 49 } 50 return 0; 51 }
上面的代码找到可行解之后需要进行比较找出最优解，我们可以对此进行优化，把queue改为priority_queue，让priority_queue帮我们完成这个工作，使得找到的第一个可行解便是满足题意的最优解。
1 #include <cstdio> 2 #include <queue> 3 using namespace std; 4 5 struct node{ 6 long long value,place; 7 bool operator < (const node& b)const 8 { 9 return value>b.value; 10 } 11 }; 12 13 void bfs(long long n) 14 { 15 node tmp; 16 tmp.value = 0; 17 tmp.place = 1; 18 priority_queue<node> q; 19 q.push(tmp); 20 while(!q.empty()){ 21 tmp = q.top(); 22 q.pop(); 23 if(tmp.value*tmp.value%tmp.place == n){ 24 printf("%I64d ",tmp.value); 25 return ; 26 } 27 node now; 28 now.place = tmp.place*10; 29 for(int i = 0; i<10; ++i){ 30 now.value = tmp.value+i*tmp.place; 31 if(now.value*now.value%now.place == n%now.place){ 32 q.push(now); 33 } 34 } 35 } 36 printf("None "); 37 } 38 39 int main() 40 { 41 int t; 42 scanf("%d",&t); 43 while(t--){ 44 long long n; 45 scanf("%I64d",&n); 46 bfs(n); 47 } 48 return 0; 49 }