xjoi奋斗群群赛12

群赛地址:

https://vjudge.net/contest/186412#problem

A - Jzzhu and Children

 CodeForces - 450A 

题意:一个小孩序列,为每个小孩需要的糖数.一人从第一个小孩开始给糖,每次给k个.求最后一个拿糖的小孩编号.

题解:

#include<iostream>using namespace std;int main(){int n,m,a[110],done[110]={0},last;cin>>n>>m;for(int i=1;i<=n;i++)cin>>a[i];bool b=1;while(b){b=0;for(int i=1;i<=n;i++){if(!done[i]){last=i;b=1;a[i]-=m;if(a[i]<=0)done[i]=1;}}}cout<<last;}

B - Jzzhu and Sequences

 CodeForces - 450B 

题意:一个序列,a[i]=a[i-1]-a[i-2],求数列第n项模1e9+7.

题解:

#include<iostream>using namespace std;const int M=1e9+7;int main(){long long a,b,n;cin>>a>>b>>n;n%=6;switch(n){case 1:cout<<(a+M)%M;break;case 2:cout<<(b+M)%M;break;case 3:cout<<(b-a+2*M)%M;break;case 4:cout<<(-a+M)%M;break;case 5:cout<<(-b+M)%M;break;case 0:cout<<(a-b+2*M)%M;break;}}

C - Jzzhu and Chocolate CodeForces - 450C

题意:把一块矩形横着竖着切k刀,使切出的最小矩形最大,输出其值.

思路:要么全切一边,要么一边切完了切另一边.

题解:

#include<iostream>#include<algorithm>using namespace std;long long n,m,k,ans;int main(){cin>>n>>m>>k;if(n<m)swap(n,m);if(k<n)ans=max(n*(m/(k+1)),m*(n/(k+1)));else ans=max(n/(k-m+2),m/(k-n+2));if(ans==0)ans=-1;cout<<ans;}

D - Jzzhu and Cities

 CodeForces - 450D 

题意:一张无向连通图,0点是首都,有k条铁路连着0与其他点,求最多能撤去的铁路数使首都到各地的距离不变.

思路:dijkstra优先队列优化,当铁路的边被替换时,ans++;

失败的代码:

#include<stdio.h>#include<string.h>#include<vector>#include<queue>using namespace std;typedef pair<int,int> pii;const int maxn=100010;const int inf=1e9+7;struct Dijkstra{int n,m;vector<pii>G[maxn];bool done[maxn];int d[maxn];void init(int n,int m){this->n=n;this->m=m;for(int i=0;i<n;i++) G[i].clear();}void AddEdge(int from,int to,int dist){G[from].push_back(pii{dist,to});G[to].push_back(pii{dist,from});}void dijkstra(int s){priority_queue<pii,vector<pii>,greater<pii> >q;for(int i=1;i<=n;i++) d[i]=inf;d[s]=0;memset(done,0,sizeof(done));q.push(pii{0,s});while(!q.empty()){pii x=q.top();q.pop();int &u=x.second;//printf("top:pii(%d %d) {\n",x.first,x.second);if(done[u])continue;done[u]=true;for(int i=0;i<G[u].size();i++){int &e=G[u][i].second;//printf("pii:(%d %d)",G[u][i].first,e);if(done[e])continue;if(d[e]>d[u]+G[u][i].first){d[e]=d[u]+G[u][i].first;q.push(G[u][i]);//printf("->in");}//printf("\n");}//printf("}\n");}}};int main(){Dijkstra DJ;int n,m,k,a,b,c,ans=0;scanf("%d%d%d",&n,&m,&k);DJ.init(n,m);for(int i=1;i<=m;i++){scanf("%d%d%d",&a,&b,&c);DJ.AddEdge(a,b,c);DJ.AddEdge(b,a,c);}DJ.dijkstra(1);//for(int i=1;i<=DJ.n;i++){//printf("%d ",DJ.d[i]);//}for(int i=1;i<=k;i++){scanf("%d%d",&a,&b);if(DJ.d[a]<=b)ans++;}printf("%d",ans);}

题解:

#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <algorithm>using namespace std;typedef long long  ll;typedef pair<int,int> pii;const int MAXV=100005;const ll INF=0x3f3f3f3f3f3f3f;int n,m,k;ll dt[MAXV];int  vis[MAXV];int  p[MAXV];vector<pii>g[MAXV];void init(){  scanf("%d%d%d",&n,&m,&k);  for(int i=0;i<m;i++){            int s,t,w;            scanf("%d%d%d",&s,&t,&w);            g[s].push_back(make_pair(t,w));            g[t].push_back(make_pair(s,w));}}int spfa(){    for(int  i=1;i<=n;i++) dt[i]=INF;    memset(p, 0, sizeof(p));queue<int>q;dt[1]=0;vis[1]=1;q.push(1);int ans=0;for(int j=0;j<k;j++){int s,d;scanf("%d%d",&s,&d);if(dt[s]>d){if(dt[s]!=INF) ans++;dt[s]=d;p[s]=1;  if(vis[s]==0){ vis[s]=1; q.push(s);}}else ans++;}    ll v,u,x;    while(!q.empty()){        v=q.front();q.pop();        vis[v]=0;        for(int  i=0;i<g[v].size();i++){            u=g[v][i].first;            x=g[v][i].second;            if(dt[v]+x<=dt[u]&&p[u]){p[u]=0;ans++;}            if(dt[v]+x<dt[u]){                dt[u]=dt[v]+x;                if(!vis[u]){                    vis[u]=1;                    q.push(u);                }            }        }    }return ans;}int main(){        init();printf("%d\n",spfa());return 0;}




E - Jzzhu and Apples
 CodeForces - 450E 

题意:一个1~n的序列,求其中最多能组成的(最大公约数>1的)数对数.

思路:从n/2的质数开始操作,统计他的倍数,倍数为偶数个则全部删去,为奇数个则保留它的2倍,其他删去.

失败的代码:

题解:

#include <stdio.h>#include <algorithm>#include <vector>#include <set>using namespace std;int n;bool p[100010], u[100010];vector<int> v;set< pair<int, int> > ans;int main() {    scanf("%d", &n);    for (int i = 2; i <= n; i++) {        if (p[i])            continue;        for (__int64 j = 1LL * i * i; j <= n; j += i)            p[j] = 1;    }    for (int i = 3; 2 * i <= n; i++) {        if (p[i])            continue;        for (int j = i; j <= n; j += i) {            if (!u[j])                v.push_back(j);        }        if (v.size() % 2)            swap(*v.rbegin(), v[1]);        for (int j = 0; j + 1 < v.size(); j += 2) {            ans.insert(make_pair(v[j], v[j + 1]));            u[v[j]] = u[v[j + 1]] = 1;        }        v.clear();    }    for (int i = 2, j = n; 1;) {        while (i < j && (u[i] || i % 2))            i++;        while (j > i && (u[j] || j % 2))            j--;        if (i >= j)            break;        ans.insert(make_pair(i++, j--));    }            printf("%d\n", ans.size());    for (set< pair<int, int> >::iterator i = ans.begin(); i != ans.end(); i++)        printf("%d %d\n", i->first, i->second);}