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

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