夏令营模拟赛 数论题

求a1x1+a2x2+a3x3+……+anxn = dy中的最小正整数解（是让y的值尽可能的小的正整数）

用图论方法最短路来做，从零点开始，记录d的剩余系中每个点已求得的多项式的最小值，最后0的情况整除d就是最优解（注意a=0的情况）

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#define ll long long
using namespace std;
ll n,d,dist[40005];
int vis[40005],flag = 0;
ll a[40005];
void spfa(){
    flag++;
    memset(dist,0,sizeof(dist));
    queue<int> q;
    q.push(0);
    ll now,nxt,step;
    while(!q.empty()){
        now = q.front();
        q.pop();
        vis[now] = 0;
        for(int i = 1;i <= n;i++){
            if(a[i] == 0) continue;
            step = dist[now] + a[i];
            nxt = now + a[i];
            if(nxt >= d){
                nxt %= d;
            }
            if(dist[nxt] > step || dist[nxt] == 0){
                dist[nxt] = step;
                if(vis[nxt] != flag){
                    vis[nxt] = flag;
                    q.push(nxt);
                }
            }
        }
    }
}
int main(){
    freopen("math.in","r",stdin);
    freopen("math.out","w",stdout);
    while(scanf("%d%d",&n,&d)){
        if(!n && !d) break;
        for(int i = 1;i <= n;i++){
            scanf("%d",&a[i]);
        }
        spfa();
        if(!(dist[0]/d))cout<<1<<endl;
        else cout<<dist[0]/d<<endl;
    }
    return 0;
}