牛客练习赛76

牛客练习赛76

A 校园活动

枚举每组的和即可

#include 

//#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9 + 7;
const int maxn = 1e3 + 10;

int a[maxn];

void solve() {
     
    int n;
    string s;
    cin>>n>>s;
    int sum=0;
    for (int i = 1; i <=n; ++i) {
     
        a[i]=s[i-1]-'0';
        sum+=a[i];
    }
    if(sum==0){
     
        cout<<n;
        return;
    }
    for (int j = 1; j <sum ; ++j) {
     
        if(sum%j==0){
     
            int now=0;
            int flag=1;
            int count=0;
            for (int i = 1; i <=n; ++i) {
     
                now+=a[i];
                if(now==j) now=0,count++;
                else if(now>j){
     
                    flag=0;
                    break;
                }
            }
            if (flag) {
     
                cout<<count<<"\n";
                return;
            }
        }
    }
    cout<<"-1\n";
}

signed main() {
     
//    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
//    cin >> _;
    while (_--) {
     
        solve();
    }
    return 0;
}

C CG的通关秘籍

#include 

//#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9+7;
const int maxn = 2e5 + 10;

int fastpow(int a,int n){
     
    int temp=a;
    int res=1;
    while (n){
     
        if(n&1) res=res*temp%mod;
        temp=temp*temp%mod;
        n>>=1;
    }
    return res;
}

void solve() {
     
    int n,m;
    cin>>n>>m;
    int gx=m*(3*m-3)/2%mod*(n-1)%mod;
    cout<<gx*fastpow(m,n-2)%mod<<"\n";
}

signed main() {
     
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--) {
     
        solve();
    }
    return 0;
}

E 牛牛数数

线性基，这是一种能够求给定序列异或第 k小的东西
二分答案 ans, 将线性基里面第 ans 小的元素与给定 x 作比较，大于则符合条件，由此二分出最小的符合条件的 ans，那么答案就是 sum - ans + 1, sum 是线性基的集合大小。

#include 

#define int long long
using namespace std;
const int maxn = 1e5 + 5;
const int mod = 77797;
const int MAXL = 63;

struct LinearBasis {
     
    long long a[MAXL + 1];
    int cnt;

    LinearBasis() {
     
        std::fill(a, a + MAXL + 1, 0);
    }

    LinearBasis(long long *x, int n) {
     
        build(x, n);
    }

    int insert(long long t) {
     
        for (int j = MAXL; j >= 0; j--) {
     
            if (!t) return 0;
            if (!(t & (1ll << j))) continue;

            if (a[j]) t ^= a[j];
            else {
     
                for (int k = 0; k < j; k++) if (t & (1ll << k)) t ^= a[k];
                for (int k = j + 1; k <= MAXL; k++) if (a[k] & (1ll << j)) a[k] ^= t;
                a[j] = t;
                return 1;
            }
        }
        return 0;
    }

    // 数组 x 表示集合 S，下标范围 [1...n]
    void build(long long *x, int n) {
     
        std::fill(a, a + MAXL + 1, 0);
        for (int i = 1; i <= n; i++) {
     
            cnt+=insert(x[i]);
        }
    }

    long long queryMax(int k) {
     
        long long res = 0;
        for (int i = 0; i <= MAXL; i++)
            if ((k>>i)&1)
                res ^= a[i];
        return res;
    }
}lb;

int b[maxn];
void solve() {
     
    int k, n;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
     
        cin>>b[i];
    }
    lb.build(b,n);
    for (int i = 0, j = 0; i <= 62; ++i)
        if (lb.a[i])
            lb.a[j++] = lb.a[i];
    int l = 0, r = (1ll << lb.cnt) - 1;
    while (l < r) {
     
        int mid = l + r + 1 >> 1;
        if (lb.queryMax(mid) > k) r = mid - 1;
        else l = mid;
    }
    cout << (1ll << lb.cnt) - l - 1;
}

signed main() {
     
    int _ = 1;
    //cin>>_;
    while (_--) {
     
        solve();
    }
    return 0;
}

F phi and phi

莫比乌斯反演，差分

欧拉函数

#include 

//#pragma GCC optimize(2)
#define int long long
using namespace std;
//const int mod = 998244353;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 10;

int vis[maxn*3];
int prime[maxn*3];
int phi[maxn*3];
int ans[maxn*3];
int num;

void init(){
     
    phi[1]=1;
    num=0;
    for (int i = 2; i < maxn; ++i) {
     
        if (!vis[i]){
     
            prime[++num]=i;
            phi[i]=i-1;
        }
        for (int j = 1; j <=num&&i*prime[j]<maxn; ++j) {
     
            vis[i*prime[j]]=1;
            if (i%prime[j]==0){
     
                phi[i*prime[j]]=phi[i]*prime[j];
                break;
            }else phi[i*prime[j]]=phi[i]*phi[prime[j]];
        }
    }
}

void solve() {
     
    init();
    int n;
    cin>>n;
    for (int T = 1; T <=n; ++T) {
     
        int sum=0;
        for (int i = 1; i <=n/T; ++i) {
     
            sum=(sum+phi[i*T])%mod;
            ans[i*T]=(ans[i*T]+phi[T]*sum%mod*sum%mod+mod)%mod;
            ans[(i+1)*T]=(ans[(i+1)*T]-phi[T]*sum%mod*sum%mod+mod)%mod;
        }
    }
    for (int i = 1; i <=n; ++i) {
     
        ans[i]=((ans[i]+ans[i-1])%mod+mod)%mod;
        cout<<ans[i]<<"\n";
    }
}

signed main() {
     
//    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
//    cin >> _;
    while (_--) {
     
        solve();
    }
    return 0;
}