AtCoder Beginner Contest 159 题解

这次的题目挺简单的awa

A

#include<bits/stdc++.h>
using namespace std;

int main(){
	int a, b; cin>>a>>b;
	cout<<a*(a-1)/2+b*(b-1)/2;
	
	return 0;
}

B

#include<bits/stdc++.h>
using namespace std;

bool ok(string s){
	string t=s;
	reverse(s.begin(), s.end());
	return t==s;
}

int main(){
	string s; cin>>s;
	if(!ok(s)){
		puts("No");
		return 0;
	}
	int n=s.size();
	s=' '+s;
	
	if(ok(s.substr(1, (n-1)/2)) && ok(s.substr((n+3)/2))) puts("Yes");
	else puts("No");
	
	return 0;
}

C

#include<bits/stdc++.h>
using namespace std;

int main(){
	double x; cin>>x; x/=3;
	printf("%.10lf", x*x*x);
	return 0;
}

D

维护一个桶和相关信息即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=2e5+5;

int n;
int w[N];

int buc[N];

ll cal(int x){
	return 1LL*x*(x-1)/2;
}

int main(){
	read(n);
	rep(i,1,n) read(w[i]), buc[w[i]]++;
	
	ll cur=0;
	rep(i,1,n) cur+=cal(buc[i]);
	 
	int pre=0;
	rep(i,1,n){
		int v=w[i];
		cur-=buc[w[i]]-1;
		buc[w[i]]--;
		cur+=buc[pre];
		buc[pre]++;
		pre=w[i];
		cout<<cur<<endl;
	}
    return 0;
}

E

注意到行数很小，所以可以直接枚举，然后大力贪心即可。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=1005;

int n, m, k;

int w[15][N], s[15][N];

int cal(int a, int b, int c, int d){
	if(!a && !b) return s[c][d];
	if(!a) return s[c][d]-s[c][b-1];
	if(!b) return s[c][d]-s[a-1][d];
	return s[c][d]-s[a-1][d]-s[c][b-1]+s[a-1][b-1];
}

int p[15], tot;

int main(){
	read(n), read(m), read(k);
	rep(i,0,n-1) rep(j,0,m-1){
		char t; cin>>t;
		s[i][j]=w[i][j]=t=='1';
		if(i) s[i][j]+=s[i-1][j];
		if(j) s[i][j]+=s[i][j-1];
		if(i && j) s[i][j]-=s[i-1][j-1];
	}	
	
	int res=INF;
	rep(i,0,(1<<n-1)-1){
		int tmp=0;
		tot=0;
		
		p[0]=-1;
		rep(j,0,n-1) if(i>>j&1) p[++tot]=j;
		p[++tot]=n-1;
		tmp=tot-1;
		// debug(i);
		// rep(i,1,tot) debug(p[i]);
		
		int last=0; // last column
		
		bool ok=true;
		rep(c,0,m-1){
			int cur=0;
			rep(x,1,tot) cur=max(cur, cal(p[x-1]+1, last, p[x], c))/*, debug(last)*/;
			// debug(cur);
			if(cur>k && last==c){
				ok=false;
				break;
			}
			if(cur>k){
				last=c;
				tmp++;
			}
		}
		
		if(ok) res=min(res, tmp);
	}
	// rep(i,0,n-1){
		// rep(j,0,m-1) cerr<<cal(i, j, i, j)<<' ';
		// cerr<<endl;
	// }
	cout<<res<<endl;
	
    return 0;
}

F

(f[x]) 表示和为 (x) 左端选择方案个数。

#pragma GCC optimize("O3")
#include<bits/stdc++.h>
using namespace std;
#define endl '
'
#define debug(x) cerr << #x << ": " << x << endl
#define pb(a) push_back(a)
#define set0(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define dwn(i,a,b) for(int i=(a);i>=(b);i--)
#define ceil(a,b) (a+(b-1))/b
#define INF 0x3f3f3f3f
#define ll_INF 0x7f7f7f7f7f7f7f7f
typedef long long ll;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;

#define int long long

inline void read(int &x) {
    int s=0;x=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {if(ch=='-')x=-1;ch=getchar();}
    while(ch>='0'&&ch<='9') s=(s<<3)+(s<<1)+ch-'0',ch=getchar();
    x*=s;
}

const int N=3030, mod=998244353;

int n, s;
int w[N], f[N];

signed main(){
	read(n), read(s);
	rep(i,1,n) read(w[i]);
	
	int res=0;
	rep(i,1,n){
		if(w[i]>s) continue;
		else{
			dwn(j,s,w[i]+1) f[j]=(f[j-w[i]]+f[j])%mod;
			f[w[i]]=(f[w[i]]+i)%mod;
			res=(res+f[s]*(n-i+1)%mod)%mod;
			f[s]=0;
		}
	}
	
	cout<<res<<endl;
	
    return 0;
}