A.数字权重

思路:

化简一下方程:

暴力有多少对a[n]和a[1]满足,中间[2,n-1]位10^(n-2) 快速幂

 

B.毒瘤xor

对于区间[L,R]的数,对第i位,若0的总数 >  1的总数 则, x[i]=1 ；否则为0

 

C.硬币游戏

模拟贪心,开4个set

 

D.粉樱花之恋

 

lgj :  , 矩阵快速幂求一下第n+1 +2项-1

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll MOD= 998244353;
 
struct Matrix {
    long long a[2][2];
    Matrix() {
        memset(a, 0, sizeof(a));
    }
    Matrix operator * (const Matrix y) {
        Matrix ans;
        for(int i = 0; i <= 1; i++)
            for(int j = 0; j <= 1; j++)
                for(int k = 0; k <= 1; k++)
                    ans.a[i][j] += a[i][k]*y.a[k][j],ans.a[i][j]%=MOD;
        for(int i = 0; i <= 1; i++)
            for(int j = 0; j <= 1; j++)
                ans.a[i][j] %= MOD;
        return ans;
    }
    void operator = (const Matrix b) {
        for(int i = 0; i <= 1; i++)
            for(int j = 0; j <= 1; j++)
                a[i][j] = b.a[i][j];
    }
};
 
ll solve(long long x) {
    Matrix ans, trs;
    ans.a[0][0] = ans.a[1][1] = 1;
    trs.a[0][0] = trs.a[1][0] = trs.a[0][1] = 1;
    while(x) {
        if(x&1)
            ans = ans*trs;
        trs = trs*trs;
        x >>= 1;
    }
    return ans.a[0][0]%MOD;
}
 
int main() {
    ll n;
    cin>>n;
    cout << ((solve(n+2)-1)%MOD+MOD)%MOD << endl;
    return 0;
}

E.符合条件的整数

import java.io.*;
import java.util.*;
import java.math.*;
  
  
public class Main {
public static void main(String[] args) {
        Scanner cin = new Scanner(new BufferedInputStream(System.in));
        int n=cin.nextInt();
        int m=cin.nextInt();
//        System.out.println(n+" "+m);
        BigInteger a=BigInteger.valueOf(1);
        BigInteger b=BigInteger.valueOf(1);
        for(int i=1;i<=n;i++)   a=a.multiply(BigInteger.valueOf(2));
        for(int i=1;i<=m;i++)   b=b.multiply(BigInteger.valueOf(2));
//        System.out.println(b+" "+a);
        a=a.subtract(BigInteger.valueOf(2));
        b=b.subtract(BigInteger.valueOf(2));
        a=a.divide(BigInteger.valueOf(7));
        if(a.compareTo(BigInteger.valueOf(0)) == 0 )  a=a.subtract(BigInteger.valueOf(1));
        b=b.divide(BigInteger.valueOf(7));
        System.out.println(b.subtract(a));
    }
}

F.可爱即正义

思路:最弱智的题

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+6;
const ll MOD=1e9+7;
const int Seed=1e7+7;
template <class T>
bool sf(T &ret){ //Faster Input
    char c; int sgn; T bit=0.1;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&c!='.'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
    if(c==' '||c=='\n'){ ret*=sgn; return 1; }
    while(c=getchar(),c>='0'&&c<='9') ret+=(c-'0')*bit,bit/=10;
    ret*=sgn;
    return 1;
}
string t="suqingnianloveskirito",s;
set<int> st;
int check(){
    int n=(int)s.size();
    bool f=false;
    int cnt=0;
//    cout <<0<<"~"<<n+1-(int)t.size()<<endl;
    for(int i=0;i<=n+1-(int)t.size();++i){
        if(s.substr(i,(int)t.size()) == t){
            ++cnt;
            st.insert(i);
//            cout <<"GO " << endl;
        }
    }
    return cnt;
}
int main(){
    cin >> s;
    int cnt=check();
//    cout << cnt << endl;
//    cout <<st.size() << endl;
    if(cnt==0)  cout <<"Yes"<<endl<<"1 2"<<endl;
    else if(cnt==1) cout <<"Yes"<<endl<<*st.begin()+1<<" "<<*st.begin()+2<<endl;
    else if(cnt==2) cout <<"Yes"<<endl<<*st.begin()+1<<" "<<*st.rbegin()+2<<endl;
    else    cout <<"No"<<endl;

    return 0;
}