Kuma Rider久违的第二场训练，这场很水，又在vj的榜单上看到第一场的大哥了，2小时ak，大哥牛啤！

A.水

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const double eps = 1e-;

int T;

char s[];

int main(void)

{

    scanf("%d", &T);

    getchar();

    while (T--)

    {

        gets(s);

        int len = strlen(s);

        int p;

        char luo[];

for(int i=;i<=;i++)luo[i]=;

        bool fu = false;

        int num1 = ;

        for (p = ; p < len; p++)

        {

            //printf("%d  %c\n", p, s[p]);

            if (s[p] == '-')fu = true;

            else if (s[p] >= ''&&s[p] <= '')

            {

                num1 *= ;

                num1 += s[p] - '';

            }

            else if (s[p] == ' ')continue;

            else

            {

                int tot = ;

                while (s[p] != ' ')

                {

                    luo[tot++] = s[p]; p++;

                }

                break;

            }

            //printf("%d  %c\n", p, s[p]);

        }

        if (fu)num1 = -num1;

        int num2 = ;

        fu = false;

        for (p = p + ; p < len; p++)

        {

            if (s[p] == '-')fu = true;

            else if (s[p] >= ''&&s[p] <= '')

            {

                num2 *= ;

                num2 += s[p] - '';

            }

        }

        if (fu)num2 = -num2;

        if (luo[] == '<')

        {

            if (luo[] == '=')

            {

                if (num1 <= num2)printf("true\n");

                else printf("false\n");

            }

            else

            {

                if (num1 < num2)printf("true\n");

                else printf("false\n");

            }

        }

        else if (luo[] == '>')

        {

            if (luo[] == '=')

            {

                if (num1 >= num2)printf("true\n");

                else printf("false\n");

            }

            else

            {

                if (num1 > num2)printf("true\n");

                else printf("false\n");

            }

        }

        else if (luo[] == '!')

        {

            if (num1 != num2)printf("true\n");

            else printf("false\n");

        }

        else

        {

            if (num1 == num2)printf("true\n");

            else printf("false\n");

        }

    }

}

B.水

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const double eps = 1e-;

int main(void) {

    int t;

    scanf("%d", &t);

    while (t--) {

        int a;

        int sum = ;

        int fla = ;

        for (int i = ; i < ; i++) {

            scanf("%d", &a);

            if (a <= ) fla = ;

            sum += a;

        }

        if (fla || sum != ) printf("NO\n");

        else printf("YES\n");

    }

    return ;

}

C.背包

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const double eps = 1e-;

const int inf = 0x3f3f3f3f;

int a[];

int dp[];

int sum;

int main(void) {

    int t;

    scanf("%d", &t);

    while (t--)

    {

        int K, m, n;

        sum = ;

        scanf("%d%d%d", &n, &m, &K);

        for (int i =; i <= K; i++)

        {

            scanf("%d", &a[i]);

            sum += a[i];

        }

        int sh = n - sum;

        if (m <= sh)printf("0\n");

        else

        {

            int need = m - sh;

            memset(dp, inf, sizeof(dp));

            dp[] = ;

            for (int i = ; i <= K; i++)

            {

                for (int j = sum; j >= a[i]; j--)

                {

                    dp[j] = min(dp[j - a[i]] + , dp[j]);

                }

            }

            int ans = inf;

            for (int i = need; i <= sum; i++)

            {

                ans = min(ans, dp[i]);

            }

            printf("%d\n", ans);

        }

    }

}

D.判断数独对不对（队友没有用函数，暴力得就很真实）

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const double eps = 1e-;

const int inf = 0x3f3f3f3f;

//

//int a[105];

//int dp[33768];

//int sum;

//int main(void) {

//    int t;

//    scanf("%d", &t);

//    while (t--)

//    {

//        int K, m, n;

//        sum = 0;

//        scanf("%d%d%d", &n, &m, &K);

//        for (int i =1; i <= K; i++)

//        {

//            scanf("%d", &a[i]);

//            sum += a[i];

//        }

//        int sh = n - sum;

//        if (m <= sh)printf("0\n");

//        else

//        {

//            int need = m - sh;

//            memset(dp, inf, sizeof(dp));

//            dp[0] = 0;

//            for (int i = 1; i <= K; i++)

//            {

//                for (int j = need; j >= 0; j--)

//                {

//                    dp[j] = min(dp[j - a[i]] + 1, dp[j]);

//                }

//            }

//            printf("%d\n", dp[need]);

//

//        }

//

//    }

//}

//

char ch[][];

int vis[];

int main(void) {

    int t;

    scanf("%d", &t);

    while (t--) {

        getchar();

        int fla = ;

        for (int i = ; i <= ; i++) {

            scanf("%s", ch[i]+);

        }

        for (int i = ; i <= ; i++) {

            memset(vis, , sizeof(vis));

            int sum = ;

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

            if (sum != ) fla = ;

        }

        for (int i = ; i <= ; i++) {

            memset(vis, , sizeof(vis));

            int sum = ;

            for (int j = ; j <= ; j++) {

                if (!vis[ch[j][i] - '']) {

                    vis[ch[j][i] - ''] = ;

                    sum++;

                }

            }

            if (sum != ) fla = ;

        }

        int sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <=; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        sum = ;

        memset(vis, , sizeof(vis));

        for (int i = ; i <= ; i++) {

            for (int j = ; j <= ; j++) {

                if (!vis[ch[i][j] - '']) {

                    vis[ch[i][j] - ''] = ;

                    sum++;

                }

            }

        }

        if (sum != ) fla = ;

        if (fla) printf("Invalid\n");

        else {

            printf("Valid\n");

        }

    }

}

E.树状数组

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const int maxn = ;

int c[maxn], a[maxn];

int n;

int lowbit(int x) {

    return x&(-x);

}

void update_onepos(int pos, int x) {

    while (pos <= n) {

        c[pos] += x;

        pos += lowbit(pos);

    }

}

int getsum_onepos(int pos) {

    int sum = ;

    while (pos > ) {

        sum += c[pos];

        pos -= lowbit(pos);

    }

    return sum;

}

int getsum_range(int x1, int x2) {

    return getsum_onepos(x2) - getsum_onepos(x1-);

}

int ax[];

int main(void) {

    int t;

    scanf("%d", &t);

    while (t--) {

        int nn;

        n = -;

        scanf("%d", &nn);

        memset(c, , sizeof(c));

        for (int i = ; i < nn; i++) {

            scanf("%d", &ax[i]);

            n = max(ax[i], n);

        }

        n += ;

        for (int i = ; i < nn; i++) {

            update_onepos(ax[i], );

        }

        int sum = ;

        for (int i = ; i < nn; i++) {

            sum += getsum_range(max(, ax[i] - ), ax[i])-;

            sum += getsum_range(ax[i] + , ax[i] + );

        }

        printf("%d\n", sum / );

    }

    return ;

}

F.并查集，最后统计根节点的问号数量

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

typedef long long LL;

const int maxv = ;

const LL mod = 1e9 + ;

int T, n, m;

char s[maxv];

bool hav ;

struct Node

{

    int pos;

    int fa;

    char ch;

}tree[];

int find(int nx)

{

    if (tree[nx].fa == nx)return nx;

    int ff = find(tree[nx].fa);

    tree[nx].ch = tree[ff].ch;

    return tree[nx].fa=ff;

}

bool vis[];

void init()

{

    hav = false;

    scanf("%d %d", &n, &m);

    scanf("%s", s + );

    for (int i = ; i <= n; i++)

    {

        if (s[i] == '?')hav = true;

        tree[i].fa = i;

        tree[i].ch = s[i];

        tree[i].pos = i;

        vis[i] = false;

    }

}

int main(void)

{

    scanf("%d", &T);

    while (T--)

    {

        init();

        bool can = true;

        for (int i = ; i <= m; i++)

        {

            int u, v;

            scanf("%d %d", &u, &v);

            int fx = find(u), fy = find(v);

            if (fx != fy)

            {

                //printf("%d %d\n", fx, fy);

                if (tree[fx].ch!='?'&&tree[fy].ch!='?')

                {

                    if (tree[fx].ch != tree[fy].ch)can = false;

                }

                else if (tree[fx].ch == '?'&&tree[fy].ch != '?')

                {

                    tree[fx].fa = fy;

                    tree[fx].ch = tree[fy].ch;

                }

                else if (tree[fx].ch != '?'&&tree[fy].ch == '?')

                {

                    tree[fy].fa = fx;

                    tree[fy].ch = tree[fx].ch;

                }

                else

                {

                    tree[fx].fa = fy;

                    tree[fx].ch = tree[fy].ch;

                }

            }

        }

        int l = , r = n;

        while (l <= r)

        {

            int fx = find(l), fy = find(r);

            if (fx != fy)

            {

                //printf("%d %d\n", fx, fy);

                if (tree[fx].ch != '?'&&tree[fy].ch != '?')

                {

                    if (tree[fx].ch != tree[fy].ch)can = false;

                }

                else if (tree[fx].ch == '?'&&tree[fy].ch != '?')

                {

                    tree[fx].fa = fy;

                    tree[fx].ch = tree[fy].ch;

                }

                else if (tree[fx].ch != '?'&&tree[fy].ch == '?')

                {

                    tree[fy].fa = fx;

                    tree[fy].ch = tree[fx].ch;

                }

                else

                {

                    tree[fx].fa = fy;

                    tree[fx].ch = tree[fy].ch;

                }

            }

            l++, r--;

        }

        if (!can){printf("0\n"); continue;}

        for (int i = ; i <= n; i++)

        {

            int fx = find(i);

            s[i] = tree[fx].ch;

        }

        //printf("%s\n", s + 1);

        LL ans = ;

        bool tag = true;

        for (int i = ; i <= n; i++)

        {

            int fx = find(i);

            if (!vis[fx])

            {

                if (tree[fx].ch == '?')

                {

                    ans *= ;

                    ans %= mod;

                }

                vis[fx] = true;

            }

        }

        if (tag)

        {

            if (hav&&ans == )printf("1\n");

            else if(!hav)printf("0\n");

            else printf("%lld\n", ans);

        }

        else printf("0\n");

    }

    return ;

}

G.状压dp

#include<cstdio>

#include<iostream>

#include<algorithm>

#include<queue>

#include<cstring>

#include<string>

#include<vector>

#include<cmath>

#include<map>

#include<functional>

#include<vector>

using namespace std;

const int maxv = ;

int T, n, m;

string s;

map<string, int>mp;

int in[maxv], w[maxv];

int dp[ << ];

int main()

{

    cin >> T;

    while (T--)

    {

        cin >> n >> m;

        memset(dp, -, sizeof(dp));

        getchar();

        for (int i = ; i <= n; i++)

        {

            in[i] = ;

            w[i] = ;

        }

        mp.clear();

        for (int i = ; i < n; i++)

        {

            getline(cin, s);

            int pos = -;

            for (int j = ; j < s.size(); j++)

            {

                if (s[j] <= ''&&s[j] >= '')

                {

                    pos = j;

                    break;

                }

            }

            string tmp = s.substr(, pos - );

            int ww = ;

            string shuzi = s.substr(pos, s.size() - pos);

            for (int j = ; j < shuzi.size(); j++)

            {

                ww = ww *  + shuzi[j] - '';

            }

            mp[tmp] = i;

            w[i] = ww;

            //cout << tmp << " " << ww << endl;

        }

        for (int i = ; i <= m; i++)

        {

            getline(cin, s);

            int pos1 = -;

            for (int i = ; i < s.size(); i++)

            {

                if (s[i] == '-')

                {

                    pos1 = i;

                    break;

                }

            }

            string t1 = s.substr(, pos1 - );

            string t2 = s.substr(pos1 + , s.size() - pos1 - );

            int id1 = mp[t1];

            int id2 = mp[t2];

            in[id2] |= ( << id1);

        }

        dp[] = ;

        for (int j = ; j < ( << n) - ; j++)

        {

            if (dp[j] < )

            {

                continue;

            }

            int num = ;

            for (int i = ; i < n; i++)

            {

                num += (j >> i & );

            }

            for (int i = ; i < n; i++)

            {

                int now = j | ( << i);

                if (j != now && (in[i] & j) == in[i])

                {

                    dp[now] = max(dp[now], dp[j] + (num + )*w[i]);

                }

            }

        }

        cout << dp[( << n) - ] << endl;

    }

    return ;

}

/*

1

3 2

Implementation 3

Dynamic Programming 10

Greedy 7

Greedy --> Dynamic Programming

Implementation --> Dynamic Programming

*/

H.模板题，待施工

gym100676 [小熊骑士限定]2015 ACM Arabella Collegiate Programming Contest的更多相关文章

  1. 边双连通缩点+树dp 2015 ACM Arabella Collegiate Programming Contest的Gym - 100676H

    http://codeforces.com/gym/100676/attachments 题目大意: 有n个城市,有m条路,每条路都有边长,如果某几个城市的路能组成一个环,那么在环中的这些城市就有传送 ...

  2. Codeforces Gym 2015 ACM Arabella Collegiate Programming Contest（二月十日训练赛）

    A(By talker): 题意分析:以a(int) op b(int)形式给出两个整数和操作符, 求两个整数是否存在操作符所给定的关系 ,有则输出true,无则输出false: 思路:由于无时间复杂 ...

  3. 2015 ACM Arabella Collegiate Programming Contest

    题目链接:https://vjudge.net/contest/154238#overview. ABCDE都是水题. F题,一开始分类讨论,结果似乎写挫了,WA了一发.果断换并查集上,A了. G题, ...

  4. 18春季训练01-3/11 2015 ACM Amman Collegiate Programming Contest

    Solved A Gym 100712A Who Is The Winner Solved B Gym 100712B Rock-Paper-Scissors Solved C Gym 100712C ...

  5. ACM Arabella Collegiate Programming Contest 2015 F. Palindrome 并查集

    题目链接:http://codeforces.com/gym/100676/attachments 题意: 给一个字符串,有一些约束条件,两个位置要相同,有一些是问号,求最后有多少种方案回文? 分析: ...

  6. ACM Arabella Collegiate Programming Contest 2015 H. Capital City 边连通分量

    题目链接:http://codeforces.com/gym/100676/attachments 题意: 有 n 个点,m 条边,图中,边强连通分量之间可以直达,即距离为 0 ,找一个点当做首都,其 ...

  7. 2017 ACM Arabella Collegiate Programming Contest(solved 11/13)

    省选考前单挑做点ACM练练细节还是很不错的嘛- 福利:http://codeforces.com/gym/101350 先来放上惨不忍睹的virtual participate成绩(中间跑去食堂吃饭于 ...

  8. 带权并查集：CF-2015 ACM Arabella Collegiate Programming Contest（F题）

    F. Palindrome Problem Description A string is palindrome if it can be read the same way in either di ...

  9. 2015 ACM Syrian Collegiate Programming Contest

    A. My Friend of Misery 计算出答案的上下界即可. 时间复杂度$O(n)$. #include<bits/stdc++.h> using namespace std; ...

随机推荐

  1. 2018 ACM-ICPC 中国大学生程序设计竞赛线上赛 H题 Rock Paper Scissors Lizard Spock.(FFT字符串匹配)

    2018 ACM-ICPC 中国大学生程序设计竞赛线上赛:https://www.jisuanke.com/contest/1227 题目链接:https://nanti.jisuanke.com/t ...

  2. c++ const static

    const作用: 1.定义常量,可以保护被修饰的东西,防止意外的修改,增强程序的健壮性. const int Max = 100; void f(const int i) { i=10;//error ...

  3. regex_iterator

    绝对不能在循环中通过regex_search获取模式在源字符串所有的实例:应该改为regex_iterator或则regex_token_iterator,一般情况下需要位一个特定的容器来指定一个尾迭 ...

  4. (未解决)记录一次登录&jmeter，留下的一地鸡毛

    一般的登录校验过程是这样的:客户端发起请求,拿到服务器给的“令牌”,再次发起请求,服务器验证“令牌”是否正确,从而返回给客户端是登录成功还是登录失败.然后我按照这个流程,用jmeter去模拟了登录过程 ...

  5. P3133 [USACO16JAN]无线电联系Radio Contact

    题目描述 Farmer John has lost his favorite cow bell, and Bessie the cow has agreed to help him find it! ...

  6. HDU 6318 Swaps and Inversions 思路很巧妙！！！（转换为树状数组或者归并求解逆序数）

    Swaps and Inversions Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Oth ...

  7. Mysql利用binlog日志恢复数据操作（转）

    a.开启binlog日志:1)编辑打开mysql配置文件/etc/mys.cnf[root@vm-002 ~]# vim /etc/my.cnf在[mysqld] 区块添加 log-bin=mysql ...

  8. muduo 网络库学习之路(一)

    前提介绍: 本人是一名大三学生,主要使用C++开发,兴趣是高性能的服务器方面. 网络开发离不开网络库,所以今天开始学一个新的网络库,陈老师的muduo库 我参考的书籍就是陈老师自己关于muduo而编著 ...

  9. 在centos环境下编译安装myrocksdb

    rocksdb(https://rocksdb.org.cn/)是脸书(facebook)公司开源的一个key-value存储引擎,基于leveldb开发.rocksdb使用的是LSM存储引擎,纯c+ ...

  10. python爬虫-execjs使用

    python爬虫-execjs使用 ecexjs的作用 通过python代码去执行JavaScript代码的库 execjs的安装 pip install PyExecJS execjs使用之前,得先 ...