AtCoder Beginner Contest 162 A~~D

A
水题：用字符串就很好处理

#include

using namespace std;

int main(){
	string s;
	cin>>s;
	for(int i=0;i<s.length();i++){
		if(s[i] == '7'){
			cout<<"Yes"<<endl;
			return 0;
		}
	}
	cout<<"No"<<endl;
} 

B.
水题

#include

using namespace std;

int main(){
	long long  n;
	cin>>n;
	long long  sum = 0;
	for(long long  i=1;i<=n;i++){
		if(i%3==0&&i%5==0){
			continue;
		}
		if(i%3==0){
			continue;
		}
		if(i%5==0){
			continue;
		}
		sum += i;
	}
	cout<<sum<<endl;
	return 0;
}

C
水题：暴力就行

#include

using namespace std;
typedef long long ll;

int main(){
	ll k;
	cin>>k;
	ll sum = 0;
	ll res;
	for(ll i=1;i<=k;i++){
		for(ll j=1;j<=k;j++){
			for(ll s=1;s<=k;s++){
				res = __gcd(i,j);
				res = __gcd(res,s);
				//cout<
				sum += res;
			}
		}
	}
	//cout<
	cout<<sum<<endl;
	return 0;
} 

D
思路：暴力肯定会超时，我采用二分，然后不知道怎么W了两个点

#include

using namespace std;

int  R[4005];
int  G[4005];
int B[4005];

int main(){
	int l;
	cin>>l;
	string s;
	cin>>s;
	int tot = 0;
	int res;
	int num1=0;
	int num2=0;
	int num3=0;
	for(int i=0;i<l;i++){
		if(s[i] == 'R'){
			R[num1] = i;
				num1++;
		} 
		if(s[i] == 'G'){
			G[num2] = i;
				num2++;
		} 
		if(s[i] == 'B'){
		B[num3] = i;	
		num3++;
		} 
	}
	if(num1==0||num2==0||num3==0) {
		cout<<"0"<<endl;
	}
	else{
		int num;
		int i,j;
		for(int i=0;i<=l-3;i++){
			for(int j=i+1;j<=l-2;j++){
				if(s[i]=='R'&&s[j]=='G'){
					res = (upper_bound(B,B+num3,j) - B);
					tot += num3 - res;
					if(binary_search(B,B+num3,2*j-i))  tot--;
				}
				if(s[i]=='G'&&s[j]=='R'){
					res = (upper_bound(B,B+num3,j) - B);
					tot += num3 - res;
					if(binary_search(B,B+num3,2*j-i)) tot--;
				}
				if(s[i]=='R'&&s[j]=='B'){
					res = (upper_bound(G,G+num2,j) -G);
					tot += num2 - res;
					if(binary_search(G,G+num2,2*j-i)) tot--;
				}
				if(s[i]=='B'&&s[j]=='R'){
					res = (upper_bound(G,G+num2,j) -G);
					tot += num2 - res;
					if(binary_search(G,G+num2,2*j-i)) tot--;
				}
				if(s[i]=='G'&&s[j]=='B'){
					res = (upper_bound(R,R+num1,j) -R);
					tot += num1 - res;
					if(binary_search(R,R+num1,2*j-i)) tot--;
				}
				if(s[i]=='B'&&s[j]=='G'){
					res = (upper_bound(R,R+num1,j) -R);
					tot += num1 - res;
					if(binary_search(R,R+num1,2*j-i)) tot--;
				}
			
			}
		}
		cout<<tot<<endl;
	}

	return 0;
}

AC代码

#include 
typedef long long ll;
using namespace std;
int main(void){
    int N; cin >> N;
    string S; cin >> S;
    
    ll r=0, g=0, b=0;
    for (auto c : S){
        if (c=='R') r++;
        if (c=='G') g++;
        if (c=='B') b++;
    }
    
    ll ans = r * g * b;
    for (int d = 1; d <= N; d++){
        for (int m = 1; m < N-1; m++){
            if (m-d<0||m+d>N-1) continue;
            if (S[m-d]!=S[m+d] && S[m-d]!=S[m] && S[m]!=S[m+d]) ans--;
        }
    }
    cout << ans << endl;
}