可惜了这场差点上大分。

A： 水题

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 1e3 + 5;
const int M = 2e4 + 5;
const double eps = 1e-6;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

void solve() { 
    string s;cin >> s;
    int pos = 0;
    for(int i = 0;i < s.size();++i) {
        if(s[i] < s[pos]) pos = i;
    }
    string ans = "";
    for(int i = 0;i < s.size();++i) if(i != pos) ans += s[i];

    cout << s[pos] << " " << ans << endl;

}   
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        solve();
    }
   // system("pause");
    return 0;
}

View Code

B：可以发现最终一定会陷入循环中，且因为和次数有关，循环的长度肯定<=n。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 2e3 + 5;
const int M = 2e4 + 5;
const double eps = 1e-6;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int a[N];
int b[N][N],cnt[N];
void solve() { 
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;++i) scanf("%d",&a[i]),b[0][i] = a[i];
    for(int j = 1;j <= n;++j) {
        memset(cnt,0,sizeof(cnt));
        for(int i = 1;i <= n;++i) cnt[b[j - 1][i]]++;
        for(int i = 1;i <= n;++i) b[j][i] = cnt[b[j - 1][i]];
    }
    int q;scanf("%d",&q);
    while(q--) {
        int x,k;scanf("%d %d",&x,&k);
        k = min(k,n);
        printf("%d\n",b[k][x]);
    }

}   
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        solve();
    }
    //system("pause");
    return 0;
}

View Code

C：可以发现，如果我们个数不够其实是可以操作0来凑的，但是如果要操作1，那么就一定要拿这位全是1的来。

所以说只有当所有的位的1的个数 % k == 0时，才能操作出来。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 2e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-6;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 1e18
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int a[N];
vector<int> ans;
int bit[30];
void solve() { 
    memset(bit,0,sizeof(bit));
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;++i) {
        scanf("%d",&a[i]);
        for(int j = 0;j < 30;++j) {
            int g = (a[i] >> j) & 1;
            bit[j] += g;
        }
    }
    ans.clear();
    ans.push_back(1);
    for(int k = 2;k <= n;++k) {
        int f = 0;
        for(int j = 0;j < 30;++j) {
            if(bit[j] % k != 0) f = 1;
        }   
        if(f == 0) ans.push_back(k);
    }
    for(auto v : ans) printf("%d ",v);
    printf("\n");
}   
int main() {
    int ca;scanf("%d",&ca);
    while(ca--) {
        solve();
    }
    //system("pause");
    return 0;
}

View Code

D：赛时想的差不多了，但是bug一直没找出来。

一开始的思路确实有点漏洞，一直考虑了线性的一个序列，但是它走的方式其实可以是上下上下来回跳的。

所以就不线性考虑了。首先有个很显然的结论，dp[i]的值肯定是第一次跳到的最小。

基于这个思路，考虑对bfs，然后线段树上维护跳到i的最小dp数就行，因为每个只会被操作一次，所以是nlogn的复杂度。

// Author: levil
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> pii;
const int N = 3e5 + 5;
const int M = 2e4 + 5;
const double eps = 1e-6;
const LL Mod = 1e9 + 7;
#define pi acos(-1)
#define INF 0x3f3f3f
#define dbg(ax) cout << "now this num is " << ax << endl;
inline long long ADD(long long x,long long y) {return (x + y) % Mod;}
inline long long DEC(long long x,long long y) {return (x - y + Mod) % Mod;}
inline long long MUL(long long x,long long y) {return x * y % Mod;}

int a[N],b[N],n;
pii pre[N];
LL dp[N];
queue<int> Q;
vector<int> vec;
struct Node{int L,r,f;}node[N << 2];
void Pushup(int idx) {node[idx].f = node[idx << 1].f & node[idx << 1 | 1].f;}
void build(int L,int r,int idx) {
    node[idx].L = L,node[idx].r = r;
    if(L == r) {
        if(L == n) node[idx].f = 1;
        else node[idx].f = 0;
        return ;
    }
    int mid = (L + r) >> 1;
    build(L,mid,idx << 1);
    build(mid + 1,r,idx << 1 | 1);
    Pushup(idx);
}
void update(int L,int r,int id,int idx) {
    if(node[idx].L >= L && node[idx].r <= r && node[idx].f) return ;
    if(node[idx].L == node[idx].r) {
        node[idx].f = 1;
        int ed = node[idx].L + b[node[idx].L];
        if(dp[ed] == 0 && ed != n) {
            Q.push(ed);
            dp[ed] = dp[id] + 1;
            pre[ed].first = id;
            pre[ed].second = node[idx].L;
        }
        return ;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    if(mid >= L) update(L,r,id,idx << 1);
    if(mid < r) update(L,r,id,idx << 1 | 1);
    Pushup(idx);
}
bool vis[N];
int BFS() {
    Q.push(n);
    dp[n] = 0;
    while(!Q.empty()) {
        int u = Q.front();
        Q.pop();
        if(u - a[u] <= 0) return u; 
        update(u - a[u],u,u,1);
    }
    return -1;
}
void solve() { 
    scanf("%d",&n);
    for(int i = 1;i <= n;++i) scanf("%d",&a[i]);
    for(int i = 1;i <= n;++i) scanf("%d",&b[i]);
    build(1,n,1);
    int ans = BFS();
    if(ans == -1) printf("-1\n");
    else {
        printf("%lld\n",dp[ans] + 1);
        vec.push_back(0);
        while(ans != n) {
            vec.push_back(pre[ans].second);
            ans = pre[ans].first;
        }
        reverse(vec.begin(),vec.end());
        for(auto v : vec) printf("%d ",v);
    }
}   
int main() {
    //int ca;scanf("%d",&ca);
    //while(ca--) {
        solve();
    //}
    //system("pause");
    return 0;
}
/*
5
0 0 1 4 2
0 2 0 1 0
*/

View Code