2016弱校连萌十一专场10.3 部分题解

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【比赛题目链接】https://www.bnuoj.com/v3/contest_show.php?cid=8504#info

【A】找两个数乘积是连续上升并且最大的。模拟即可。

【代码君】

////Created by just_sort 2016/10/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 1010;int a[maxn];int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        for(int i = 1; i <= n; i++) scanf("%d",&a[i]);        int ans = -1;        for(int i = 1; i <= n; i++)        {            for(int j = 1; j < i; j++){                int tmp = a[i]*a[j];                int p = tmp%10; tmp /= 10;                while(tmp)                {                    int q = tmp%10;                    if(q + 1 != p) break;                    tmp /= 10;                    p = q;                }                if(tmp == 0) ans = max(ans,a[i]*a[j]);            }        }        printf("%d\n",ans);    }    return 0;}

【B】从终点开始bfs，记录各点到终点最短路。

【代码君】

////Created by just_sort 2016/10/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int maxn = 220;const int inf = 0x3f3f3f3f;int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};struct node{int x,y;};queue<node>q;int dp[maxn][maxn];char G[maxn][maxn];int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        int ans1,ans2;        for(int i = 0; i < n; i++) scanf("%s",G[i]);        for(int i = 0; i < n; i++){            for(int j = 0; j < m; j++){                if(G[i][j] == '%'){                    q.push(node{i,j});                }            }        }        ans1 = ans2 = inf;        memset(dp,-1,sizeof(dp));        while(q.size())        {            node now = q.front();            q.pop();            if(G[now.x][now.y] == '$') ans1 = min(ans1,dp[now.x][now.y]);            if(G[now.x][now.y] == '@') ans2 = min(ans2,dp[now.x][now.y]);            for(int i = 0; i < 4; i++)            {                int dx = now.x + dir[i][0];                int dy = now.y + dir[i][1];                if(dx >= 0 && dx < n && dy >= 0 && dy < m && G[dx][dy] != '#' && dp[dx][dy] == -1){                    dp[dx][dy] = dp[now.x][now.y] + 1;                    q.push(node{dx,dy});                }            }        }        if(ans2 < ans1) printf("Yes\n");        else printf("No\n");    }    return 0;}

【C】n对括号最多需要交换n*(n+1)/2次（)))...(((类似这种摆放情况），那么找最短的满足m*(m+1)/2>=n的，然后把额外需要交换的交换完毕就是结果

【代码君】

////Created by just_sort 2016/10/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;int main(){    int n;    while(scanf("%d",&n)!=EOF)    {        int m = 1;        while(m * (m + 1) / 2 < n) m++;        string str;        str = string(m,')')+string(m,'(');        int now = m;        for(int i = 0; i < m * (m + 1) / 2 - n; i++){            swap(str[now-1],str[now]);            now--;        }        cout<<str<<endl;    }    return 0;}

【D】ACM协会中有 n 个人, 每个人有个 motivation level , 将每个人按照 motivation level 从大到小排序, 相同的按照入会时间从小到大排序. 协会里面排名在前 20 的会努力工作, 剩下80 不工作. 现在有m 个事件, 加入一个新的成员或者一个成员离开. 每次事件后输出工作状态变化的人. 保证同时只有一个人状态变化.

【解题方法】set大模拟，开两个集合，一个是 s1 装的是 20 的人的信息，另一个集合 s2 装的是剩下的信息，注意开一个结构体，这个结构体中也包括时间，然后s1 集合中是按照那个motivation level 从小到大排序的，然后 s2 是motivation level 从大到小排序的，然后就进行模拟就行了。

【代码君】

////Created by just_sort 2016/10/7//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e5+6;struct node{    string s;    int time;    int d;};bool cmp(node a,node b){    if(a.d == b.d) return a.time > b.time;    else return a.d>b.d;}struct cmp1{//从大到小排序    bool operator()(const node &a,const node &b)const{        if(a.d == b.d) return a.time > b.time;        else return a.d > b.d;    }};struct cmp2{//从小到大排序    bool operator()(const node &a,const node &b)const{        if(a.d == b.d) return a.time < b.time;        else return a.d < b.d;    }};set<node,cmp2>s1;set<node,cmp1>s2;map<string,node>mp;node a[maxn];int main(){    int n,m;    while(scanf("%d",&n)!=EOF)    {        s1.clear();        s2.clear();        mp.clear();        double tmp = 1.0*n*0.2;        int x = floor(tmp);        node xx;        for(int i = 1; i <= n; i++){            cin>>a[i].s>>a[i].d;            a[i].time = i;            mp[a[i].s] = a[i];        }        int curn = n;        sort(a+1,a+n+1,cmp);        for(int i = 1; i <= x; i++){            s1.insert(a[i]);        }        for(int i = x + 1; i <= n; i++){            s2.insert(a[i]);        }        cin>>m;        string s;        for(int i = curn+1; i <= curn + m; i++){            char op[5];            scanf("%s",op);            if(op[0] == '-')//'-'            {                cin>>s;                xx = mp[s];                if(s1.erase(xx)) x--;                s2.erase(xx);                if(x > (int)(1.0*(n-1)*0.2))                {                    x--;                    xx = *s1.begin();                    s1.erase(xx);                    s2.insert(xx);                    cout<<xx.s;                    printf(" is not working now.\n");                }                n--;                if(x < (int)(1.0*(n)*0.2))                {                    x++;                    xx = *s2.begin();                    s1.insert(xx);                    s2.erase(xx);                    cout<<xx.s;                    printf(" is working hard now.\n");                }            }            else if(op[0] == '+') //'+'            {                cin>>a[i].s>>a[i].d;                a[i].time = i;                mp[a[i].s] = a[i];                if(x < (int)(1.0*(n+1)*0.2))                {                    if(a[i].d > (*s2.begin()).d || (a[i].d == (*s2.begin()).d && a[i].time > (*s2.begin()).time))                    {                        s1.insert(a[i]);                        cout<<a[i].s;                        printf(" is working hard now.\n");                    }                    else                    {                        xx = *s2.begin();                        s2.erase(xx);                        s1.insert(xx);                        s2.insert(a[i]);                        cout<<a[i].s;                        printf(" is not working now.\n");                        cout<<xx.s;                        printf(" is working hard now.\n");                    }                    x++;                }                else                {                    if(x != 0)                    {                        xx = *s1.begin();                        if(a[i].d > xx.d || (a[i].d == xx.d && a[i].time > xx.time))                        {                            s1.erase(xx);                            s1.insert(a[i]);                            s2.insert(xx);                            cout<<a[i].s;                            printf(" is working hard now.\n");                            cout<<xx.s;                            printf(" is not working now.\n");                        }                        else{                            s2.insert(a[i]);                            cout<<a[i].s;                            printf(" is not working now.\n");                        }                    }                    else{                        xx = *s2.begin();                        if((int)(1.0*(n+1)*0.2) > 0)                        {                            if(a[i].d > xx.d || (a[i].d == xx.d && a[i].time > xx.time))                            {                                s1.insert(a[i]);                                cout<<a[i].s;                                printf(" is working hard now.\n");                            }                            else{                                s2.erase(xx);                                s2.insert(a[i]);                                s1.insert(xx);                                cout<<a[i].s;                                printf(" is not working now.\n");                                cout<<xx.s;                                printf(" is working hard now.\n");                            }                        }                        else{                            s2.insert(a[i]);                            cout<<a[i].s;                            printf(" is not working now.\n");                        }                    }                }                n++;            }        }    }    return 0;}

【E】贴个题解http://www.cnblogs.com/chen9510/p/5929542.html

【代码君】

////Created by just_sort 2016/10/3//Copyright (c) 2016 just_sort.All Rights Reserved//#include <set>#include <map>#include <queue>#include <stack>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;typedef unsigned long long ull;const int maxn = 111111 + 10;int n;vector<int> g[maxn];map<ull, int> mp;ull dfs(int u, int f, int p) {ull res = 0;for (int i = 0; i < g[u].size(); ++i) {int v = g[u][i];res += dfs(v, u, p);}res = res * p + 1;mp[res]++;return res;}int main() {scanf("%d", &n);for (int i = 0; i < n - 1; ++i) {int u, v;scanf("%d%d", &u, &v);g[u].push_back(v);}mp.clear();dfs(1, 0, 1e9 + 7);ll ans = 0;map<ull, int>::iterator it;for (it = mp.begin(); it != mp.end(); it++) {ll x = it->second;ans += x * (x - 1) / 2;}printf("%lld\n", ans);}

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