A

描述

AC代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <set>
#include <stack>
#include <iomanip>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define seteps(N) setprecision(N)
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef long long ll;
typedef pair<ll,ll> PLL;
typedef __int128 lll;
const int N=2*1010;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll one=1;
ll f[N][N];
void solve(){
    double p=3.1415926;
    int n,m;
    while(cin>>n>>m){
        rep(i,0,n+m+1){
            rep(j,0,n+m+1){
                f[i][j]=0;
            }
        }
        f[0][0]=1;
        rep(i,0,n+m+1){
            rep(j,0,n+m+1){
                if(i>=1 && i-j<=n) f[i][j]=(f[i][j]+f[i-1][j])%mod;
                if(j>=1 && j-i<=m) f[i][j]=(f[i][j]+f[i][j-1])%mod;
            }
        }
        cout<<f[n+m][n+m]<<endl;
    }
}
int main(){
    IOS;
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    //int t;cin>>t;
    //while(t--)
    solve();
    return 0;
}

B

题意描述

AC代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <set>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef long long ll;
typedef pair<ll,ll> PLL;
typedef __int128 lll;
const int N=2*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll one=1;
int a[N];
void solve(){
    int n;
    cin>>n;
    rep(i,1,n+1) cin>>a[i];
    ll ans=0;
    map<int,int> used;
    rep(i,1,n+1){
        if(!used[a[i]]){
            ans+=(i*(n-i+1));
        }else{
            ans+=(n-i+1)*(i-used[a[i]]);
        }
        used[a[i]]=i;
    }
    cout<<ans<<endl;
}
int main(){
    IOS;
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    //int t;cin>>t;
    //while(t--)
    solve();
    return 0;
}

F

AC代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <set>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef long long ll;
typedef pair<ll,ll> PLL;
typedef __int128 lll;
const int N=2*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll one=1;
int a[N];
void solve(){
    ll a,b,x,y;
    while(cin>>x>>a>>y>>b){
        lll res1=one*x*b;
        lll res2=one*a*y;
        if(res1==res2) cout<<"="<<endl;
        else if(res1<res2) cout<<"<"<<endl;
        else cout<<">"<<endl;
    }
}
int main(){
    IOS;
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    //int t;cin>>t;
    //while(t--)
    solve();
    return 0;
}

G

题意描述

思路

使用一个字符串来储存删除过后的字符串序列，使用一个变量来表示删除后的字符串下标。每次符合条件时，变量都要向前移3位，模拟这个过程即可。

AC代码

#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <cstdio>
#include <string>
#include <cstring>
#include <set>
#include <stack>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define sz(x) x.size()
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<char,char> PCC;
typedef long long ll;
typedef pair<ll,ll> PLL;
typedef __int128 lll;
const int N=2*1e5+10;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const lll one=1;
void solve(){
    string s,v;
    cin>>s;
    v=s;
    int now=0,ans=0;
    rep(i,0,sz(s)){
        now++;
        v[now]=s[i];
        if(now>=3 && v[now]==v[now-1] && v[now-2]==v[now]){
            ans++;
            now-=3;
        }
    }
    cout<<ans<<endl;
}
int main(){
    IOS;
    //freopen("test.in", "r", stdin);
    //freopen("test.out", "w", stdout);
    //int t;cin>>t;
    //while(t--)
    solve();
    return 0;
}