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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/hyfer/p/5812507.html
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