POI X Sums(同余最短路)

Sums

Memory limit: 32 MB

We are given a set of positive integers . Consider a set of non-negative integers , such that a number  belongs to  if and only if  is a sum of some elements from  (the elements may be repeated). For example, if , then sample numbers belonging to the set  are: 0 (the sum of 0 elements), 2, 4  and 12  or  or ; and the following do not belong to : 1 and 3.

Task

Write a program which:

• reads from the standard input the description of the set  and the sequence of numbers ,
• for each number  determines whether it belongs to the set ,
• writes the result to the standard output.

Input

In the first line there is one integer : the number of elements of the set , . The following  lines contain the elements of the set , one per line. In the -st line there is one positive integer , . , .

In the -nd line there is one integer , . Each of the following  lines contains one integer in the range from  to , they are respectively the numbers .

Output

The output should consist of  lines. The -th line should contain the word TAK ("yes" in Polish), if  belongs to , and it should contain the word NIE ("no") otherwise.

Example

For the input data:

325760141232

the correct result is:

TAKNIETAKTAKNIETAK

这道题的地址为：http://main.edu.pl/en/archive/oi/10/sum 

or

https://szkopul.edu.pl/problemset/problem/4CirgBfxbj9tIAS2C7DWCCd7/site/?key=statement

这是叉姐在他的网站发出来的一道题，我大致理解了叉姐说的是啥意思，自己写了一发，但是这个评测网站，，，不知道是怎样的操作，反正交了题好久好久没有评测结果。

反正思路就是，先把n个数字a_i读入，然后从0开始跑最短路。建好图之后，直接读入k个数字进行判断即可。

其中，最短路的dis[i]表示的是，到达i所需要的最小步数。这个i是大于0小于自己定的一个模数mod的。

取最小的那个a复杂度肯定最小嘛，复杂度是O(mod log mod)。

接下来读入k个数字b_i，那么我们将他对mod取模得t，比对dis[t],如果b比t小，那是无法这么早就到达的，输出NIE，否则肯定有办法到达，输出TAK。

实在不懂，手玩一下样例就很好理解了。

emmmm。。。过了过了23333

代码如下：

#include<iostream>#include<cstdio>#include<queue>#include<algorithm>#include<cstring>using namespace std;int mod = 50000;int dis[50005], a[50005];bool vst[50005];int n;void dijkstra() {memset(dis, 0x3f, sizeof(dis));priority_queue<int, vector<int>, greater<int> >q;dis[0] = 0;q.push(0);while(!q.empty()) {int u= q.top();q.pop();if(vst[u])continue;vst[u] = 1;for(int i = 0; i < n; i++) {if(dis[u] + a[i] < dis[ (u + a[i]) % mod ]) {dis[ (u + a[i]) % mod ] = dis[u] + a[i];q.push( (u + a[i]) % mod );}}}}int main() {ios_base::sync_with_stdio(0);cin >> n;for(int i = 0; i < n; i++) {cin >> a[i];} sort(a, a + n);mod = a[0];//好像直接拿个Min一路比过来也行-- dijkstra();int k, t;cin >> k;while(k--) {cin >> t;if( t < dis[ (t % mod) ] )cout << "NIE\n";elsecout << "TAK\n";}return 0;}