A.算2,3的因子个数即可

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int main(){

    Sca2(N,M);

    if(M % N){

        puts("-1");

        return ;

    }

    M /= N;

    int ans = ;

    while(!(M % )){

        M /= ; ans++;

    }

    while(!(M % )){

        M /= ; ans++;

    }

    if(M != ) ans = -;

    Pri(ans);

    return ;

}

A

B.复制成两倍长找最长连续1

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 5e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

int a[maxn];

int main(){

    Sca(N);

    for(int i = ; i <= N ; i ++) a[i] = a[i + N] = read();

    N <<= ;

    int cnt = ;

    int ans = ;

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

        if(a[i]){

            cnt++;

            ans = max(ans,cnt);

        }else{

            cnt = ;

        }

    }

    Pri(ans);

    return ;

}

B

C.设第一个数字为x,算前缀和pre可知a[i] = x + pre[i],找pre[i]最小的位置即为1的位置,可推知整个数组

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 5e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL a[maxn];

bool vis[maxn];

bool check(LL x){

    if( > x || x > N) return false;

    if(vis[x]) return false;

    vis[x] = ;

    return true;

}

int main(){

    Sca(N);

    a[] = ;

    LL Min = ;

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

        Scl(a[i]);

        a[i] += a[i - ];

        Min = min(Min,a[i]);

    }

    a[] =  - Min;

    bool flag = ;

    if(!check(a[])) flag = ;

    for(int i = ; i <= N && flag; i ++){

        a[i] += a[];

        if(!check(a[i])) flag = ;

    }

    if(!flag) puts("-1");

    else for(int i = ; i <= N ; i ++) cout << a[i] << " ";

    return ;

}

C

D.贪心,存下第一个字符串,第二个字符串的小写字母能直接匹配就匹配,不能匹配就匹配问号,最后将第二字符串的问号匹配第一个字符串落单的

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

vector<int>Q[];

vector<int>P;

char str1[maxn],str2[maxn];

vector<PII>ans;

int main(){

    Sca(N);

    scanf("%s%s",str1 + ,str2 + );

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

        if(str1[i] == '?') Q[].pb(i);

        else Q[str1[i] - 'a'].pb(i);

    }

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

        if(str2[i] == '?') P.pb(i);

        else{

            if(!Q[str2[i] - 'a'].empty()){

                ans.pb(mp(Q[str2[i] - 'a'].back(),i));

                Q[str2[i] - 'a'].pop_back();

            }else if(!Q[].empty()){

                ans.pb(mp(Q[].back(),i));

                Q[].pop_back();

            }

        }

    }

    for(int i =  ; i <  && !P.empty(); i ++){

        while(!P.empty() && !Q[i].empty()){

            ans.push_back(mp(Q[i].back(),P.back()));

            P.pop_back(); Q[i].pop_back();

        }

    }

    Pri(ans.size());

    for(int i =  ; i < ans.size(); i ++){

        printf("%d %d\n",ans[i].fi,ans[i].se);

    }

    return ;

}

D

E.求个前缀和,记录下最小前缀和和整个数组的和.

难点在于求会经过几个完整的回合才会遇到最终死的那个回合,这里用了倍增的方法处理,但我觉得显然可以用一个式子直接计算,只是因为我菜没想到

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = 2e5 + ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL H;

LL a[maxn];

LL pre[maxn];

int main(){

    Scl(H); Sca(N);

    LL Min = 1e18;

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

        Scl(a[i]);

        pre[i] = pre[i - ] + a[i];

        Min = min(Min,pre[i]);

    }

    if(H + Min >  && pre[N] >= ){

        puts("-1");

        return ;

    }

    LL ans = ;

    if(H + Min > ){

        LL x = ;

        while(pre[N] * x + H + Min > ){

            x *= ;

        }

        x /= ;

        for(LL i = x; i >  ; i /= ){

            while(pre[N] * i + H + Min > ){

                H += pre[N] * i;

                ans += i * N;

            }

        }

    }

    int cnt = ;

    while(H > ){

        ans++;

        H += a[++cnt];

        if(cnt == N) cnt = ;

    }

    Prl(ans);

    return ;

}

E

F.先求一个前缀和,很显然区间和为pre[r] - pre[l - 1],先用一个n²的操作求出所有可能的和,然后用类似邻接表的方法,以区间和为key值,区间的l,r为value值存储下来.

可以将key值hash,我这里用了map<int,pair<int,int>>Q,的方式暴力存储.

对于每个key值都寻找答案,贪心的将他们所有的区间按右端点排序就可以找到这个key值下最多的区间数目.

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL a[maxn],pre[maxn];

map<int,vector<PII>>Q;

map<int,bool>P;

bool cmp(PII a,PII b){

    return a.se < b.se;

}

int main(){

    Sca(N);

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

        Sca(a[i]);

        pre[i] = pre[i - ] + a[i];

    }

    int ans = ;

    for(int l = ; l <= N ; l ++){

        P.clear();

        for(int r = l; r <= N ; r ++){

            int sum = pre[r] - pre[l - ];

            if(P[sum]) continue;

            P[sum] = ;

            Q[sum].push_back(mp(l,r));

        }

    }

    map<int,vector<PII>>::iterator A;

    for(map<int,vector<PII>>::iterator it = Q.begin(); it != Q.end(); it++){

        vector<PII>q = (*it).se;

        sort(q.begin(),q.end(),cmp);

        int t = ;

        int cnt = ;

        for(int i =  ; i < q.size(); i ++){

            if(q[i].fi > t){

                cnt++;

                t = q[i].se;

            }

        }

        if(ans < cnt) A = it;

        ans = max(ans,cnt);

    }

    Pri(ans);

    vector<PII>q = (*A).se;

    sort(q.begin(),q.end(),cmp);

    int t = ;

    for(int i =  ; i < q.size(); i ++){

        if(q[i].fi > t){

            cout << q[i].fi << " " << q[i].se << endl;

            t = q[i].se;

        }

    }

    return ;

}

F

G.画画图就很显然的指导若x为公司数量ind[i] <= x的点必可以为好点,反之必为坏点.

用二分找出最小的满足坏点 <= k的公司数x,贪心的方法得到树边的染色.

#include <map>

#include <set>

#include <ctime>

#include <cmath>

#include <queue>

#include <stack>

#include <vector>

#include <string>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <sstream>

#include <iostream>

#include <algorithm>

#include <functional>

using namespace std;

#define For(i, x, y) for(int i=x;i<=y;i++)

#define _For(i, x, y) for(int i=x;i>=y;i--)

#define Mem(f, x) memset(f,x,sizeof(f))

#define Sca(x) scanf("%d", &x)

#define Sca2(x,y) scanf("%d%d",&x,&y)

#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)

#define Scl(x) scanf("%lld",&x);

#define Pri(x) printf("%d\n", x)

#define Prl(x) printf("%lld\n",x);

#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();

#define LL long long

#define ULL unsigned long long

#define mp make_pair

#define PII pair<int,int>

#define PIL pair<int,long long>

#define PLL pair<long long,long long>

#define pb push_back

#define fi first

#define se second

typedef vector<int> VI;

int read(){int x = ,f = ;char c = getchar();while (c<'' || c>''){if (c == '-') f = -;c = getchar();}

while (c >= ''&&c <= ''){x = x *  + c - '';c = getchar();}return x*f;}

const double eps = 1e-;

const int maxn = ;

const int INF = 0x3f3f3f3f;

const int mod = 1e9 + ;

int N,M,K;

LL a[maxn],pre[maxn];

map<int,vector<PII>>Q;

map<int,bool>P;

bool cmp(PII a,PII b){

    return a.se < b.se;

}

int main(){

    Sca(N);

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

        Sca(a[i]);

        pre[i] = pre[i - ] + a[i];

    }

    int ans = ;

    for(int l = ; l <= N ; l ++){

        P.clear();

        for(int r = l; r <= N ; r ++){

            int sum = pre[r] - pre[l - ];

            if(P[sum]) continue;

            P[sum] = ;

            Q[sum].push_back(mp(l,r));

        }

    }

    map<int,vector<PII>>::iterator A;

    for(map<int,vector<PII>>::iterator it = Q.begin(); it != Q.end(); it++){

        vector<PII>q = (*it).se;

        sort(q.begin(),q.end(),cmp);

        int t = ;

        int cnt = ;

        for(int i =  ; i < q.size(); i ++){

            if(q[i].fi > t){

                cnt++;

                t = q[i].se;

            }

        }

        if(ans < cnt) A = it;

        ans = max(ans,cnt);

    }

    Pri(ans);

    vector<PII>q = (*A).se;

    sort(q.begin(),q.end(),cmp);

    int t = ;

    for(int i =  ; i < q.size(); i ++){

        if(q[i].fi > t){

            cout << q[i].fi << " " << q[i].se << endl;

            t = q[i].se;

        }

    }

    return ;

}

G

codeforces-1141 (div3)的更多相关文章

  1. Codeforces #550 (Div3) - G.Two Merged Sequences（dp / 贪心）

    Problem  Codeforces #550 (Div3) - G.Two Merged Sequences Time Limit: 2000 mSec Problem Description T ...

  2. codeforces #579(div3)

    codeforces #579(div3) A. Circle of Students 题意: 给定一个n个学生的编号,学生编号1~n,如果他们能够在不改变顺序的情况下按编号(无论是正序还是逆序,但不 ...

  3. CodeForces 1029E div3

    题目链接 第一道场上自己做出来的E题...虽然是div3,而且是原题... 当时做完ABC,D题没有思路就去怼E了,然后发现貌似原题? 事实上就是原题... 给个原题链接... [HNOI2003]消 ...

  4. [Codeforces #615 div3]1294E Obtain a Permutation

    Before the Beginniing 本文为 Clouder 原创文章,原文链接为Click,转载时请将本段放在文章开头显眼处.如进行了二次创作,请明确标明. 由本人转载于博客园. 题意分析 C ...

  5. Codeforces #624 div3 C

    You want to perform the combo on your opponent in one popular fighting game. The combo is the string ...

  6. Twist the Permutation 数列的轮换题 Codeforces 776 div3

    这是一道比较经典的将数列中的数字轮换的题目,我们先看题干: 题干分析:先浅浅地分析一下题目是要我们干什么,我们会默认有一个已经升序排序地1~n的排列,然后我们会给定一个新排列是在原有排列的基础上进行o ...

  7. 记一次神奇的codeforces

    今天有一场codeforces的div3,时间挺合适,于是就想打.结果发现rating超过1600就不能报名.虽然shzr好久不打CF了而且很菜,但是毕竟还是到了1600的,于是和ZUTTER_一起用 ...

  8. CodeForces Round #527 (Div3) B. Teams Forming

    http://codeforces.com/contest/1092/problem/B There are nn students in a university. The number of st ...

  9. 【赛时总结】◇赛时·V◇ Codeforces Round #486 Div3

    ◇赛时·V◇ Codeforces Round #486 Div3 又是一场历史悠久的比赛,老师拉着我回来考古了……为了不抢了后面一些同学的排名,我没有做A题 ◆ 题目&解析 [B题]Subs ...

  10. Codeforces Round #535 (Div. 3) [codeforces div3 难度测评]

    hhhh感觉我真的太久没有接触过OI了 大约是前天听到JK他们约着一起刷codeforces,假期里觉得有些颓废的我忽然也心血来潮来看看题目 今天看codeforces才知道居然有div3了,感觉应该 ...

随机推荐

  1. Redis(十一):Redis的事务功能详解

    相关命令 1. MULTI 用于标记事务块的开始.Redis会将后续的命令逐个放入队列中,然后才能使用EXEC命令原子化地执行这个命令序列. 这个命令的运行格式如下所示: MULTI 这个命令的返回值 ...

  2. c/c++ 拷贝控制 构造函数的问题

    拷贝控制 构造函数的问题 问题1:下面①处的代码注释掉后,就编译不过,为什么??? 问题2:但是把②处的也注释掉后,编译就过了,为什么??? 编译错误: 001.cpp: In copy constr ...

  3. springboot项目屏蔽mq或者mongodb的监控日志输出

    最近写项目,用的是springboot,其中用到了rabbitmq和mongodb,配置完成 项目启动后,会输出如下日志: mongodb和mq的检测,会一直打印日志,这样会影响开发人员的测试. 如何 ...

  4. 【Linux基础】grep命令

    1.简介 grep是一种强大的文本搜索工具,它能使用正则表达式搜索文本,并把匹配的行打印出来. 命令格式:grep [option] pattern file 2.常用参数与举例: -e :  使用P ...

  5. 研究好vif 和vshow

    另外从源头上处理的???,怎么自己排查出错误??必须 ??https://www.jb51.net/article/124116.htm

  6. vue 路由变化页面数据不刷新问题(缓存)

    每天记录一点点,把我遇到的问题记录下来, 希望可以帮助到更多和我遇到同样问题的人. 问题描述:通过调接口,动态显示帮助页面的问题列表, 问题列表有多级,当点击的这个问题没有下一级问题的时候跳入内容页. ...

  7. android开发学习 ------- 关于getSupportFragmentManager()不可用的问题

    在Android开发中,少不了Fragment的运用. 目前在实际运用中,有v-4包下支持的Fragment以及app包下的Fragment,这两个包下的FragmentManager获取方式有点区别 ...

  8. Golang 入门系列（六）理解Go中的协程（Goroutine）

    前面讲的都是一些Go 语言的基础知识,感兴趣的朋友可以先看看之前的文章.https://www.cnblogs.com/zhangweizhong/category/1275863.html. 今天就 ...

  9. 乡下人设计模式——SOLID之六大原则

    S(Single Responsibility Principle):单一责任原则 O(Open Closed Principle):开放封闭原则 L(Liskov Substitution Prin ...

  10. 关于childNodes的删除

    在使用childNodes时,发现需要删除的元素多于1时,会出现无法全部删除的情况.谷歌以后发现,该属性返回的子节点集合是实时更新的,也就是说,在for循环中,当删除第一个子节点之后,第二次删除的是原 ...