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2022.1.24牛客训练营1补题日记

莞尔小迷糊 2022-01-26 阅读 65
算法

补题日记1

2022.1.24牛客寒假算法基础集训营1
呜呜呜我是傻逼只做出来最基础的L签到题，简直是最简单的有手就行的，其他的我有尝试过EFH，我是撒比，都写了，都过不了，于是本人下定决心决定要补题！！上传csdn以此督促自己呜呜呜。

H题

二分加前缀和
1

//来自我的撒比想法，因为n=1e6,n^2超时，时间复杂度O(N^2)
    for(int i=0;i<n;i++)cin>>a[i];
    for(int i=0;i<n;i++){
        for(int j=i;j<n;j++){
            res+=abs(a[i]+a[j]-1000);
        }
    }
    cout<<res<<endl;

2

//后生可畏cdf的hs记录次数写法,时间复杂度O(值域^2)=O(10^3^2),如果值域过大就不能这样
 	long long ans=0;
    vector<long long> hs(1010);
    for(int i=0;i<n;i++){
        if(cin>>x)hs[x]++;
    }
    for(int i=0;i<=1010;i++){
        for(int j=i+1;j<=1000;j++){
            ans+=abs(i+j-1000)*hs[i]*hs[j];
        }
        ans+=abs(i+i-1000)*(hs[i]-1)*hs[i]/2;
        ans+=abs(i+i-1000)*hs[i];
    }
    cout<<ans;

3

//来自康康子的二分加前缀和写法 时间复杂度O(N*logN) ;lowerbound查询复杂度O(logN)
#include<bits/stdc++.h>
using namespace std;
int n;
int main(){
    ios::sync_with_stdio(0);
    cin>>n;
    long long ans=0;
    vector<long long> a(n),pre(n+1);
    for(int i=0;i<n;i++)cin>>a[i];
    sort(a.begin(),a.end());
    for(int i=0;i<n;i++)pre[i+1]=pre[i]+a[i];
    
    for(int i=0;i<n;i++){
        int j=lower_bound(a.begin()+i,a.end(),1000-a[i])-a.begin();
        ans+=pre[n]-pre[j]+(n-j)*a[i]-(n-j)*1000;
        ans+=1000*(j-i)-(pre[j]-pre[i]+(j-i)*a[i]);
    }
    cout<<ans;
    return 0;
}

lowerboundupperbound
4

//网友的简短写法
const int N=1e6+5;
int n;
ll a[N],sum[N];
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        sum[a[i]]++;
    }
    ll ans=0;
    for(int i=0;i<=1000;i++){
        ans+=sum[i]*abs(i+i-1000);
        for(int j=0;j<=1000;j++)
            ans+=sum[i]*sum[j]*abs(i+j-1000);
    }
    cout<<ans/2<<endl;
}

5

//纯纯二分和前缀和
//#define IOS ios::sync_with_stdio(false),cin.tie(0);
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<char, pair<int, int> > PCII;
typedef pair<int, pair<int, int> > PIII;
const int maxn=1e6+7;
const int N = 1e5+7;
const int mod = 998244353;
const ll inf=1e18;
const double pi = 3.14159265359;
int sum[maxn];
int n;
vector<int> v,a;
int main()
{
	a.push_back(0);
	cin>>n;
	for(int i=1;i<=n;i++)
	{
		int x;
		cin>>x;
		a.push_back(x);
	}
	sort(a.begin(),a.end());
	for(int i=1;i<=n;i++) sum[i] = sum[i-1] + a[i];
	
	ll ans = 0;
	for(int i=1;i<=n;i++)
	{
		int num = 1000 - a[i];	//临界值
		int l = i,r = n;
		while(l < r)
		{
			int mid = l + r + 1>> 1;
			if(a[mid] > num) r = mid - 1;
			else l = mid;
		}
		int idx = l;
	int idx = lower_bound(i-1+a.begin(),a.end(),num) - a.begin();
	cout<<a.begin()<<endl;
//		cout<<l<<endl;
		if(idx != a.size())	//找到了(后面全是>=num的，前面全是<=num的) 
		{
			ans += ((n-idx+1)*a[i] + (sum[n] - sum[idx-1]) - (n-idx+1)*1000);
			ans += (1000*(idx-i) - (idx-i)*a[i] - (sum[idx-1] - sum[i-1]));
		}
		else
			ans += (1000*(n-i) - (n-i)*a[i] - (sum[idx-1] - sum[i-1]));
	}
	cout<<ans<<endl;
	return 0;
}
//499 500 500 501
//499 999 1499 2000

E题
在这里插入图片描述
1

//来自我的撒比想法，为什么为什么！样例都过了！！但是就是过不了！！
#include<bits/stdc++.h>
using namespace std;
long long t,res=0,x=0;
long long n,m;
int main(){
    cin>>t;
    while(t--){
        cin>>n>>m;
        if(n==m){cout<<1<<endl;continue;}
        else if(n>1&&m==1){cout<<-1<<endl;continue;}
        else if(n>m){
            x++;res+=2;
            while(n-x*m+x>m){res+=2;x++;}
            if(n-x*m+x<=m){res++;}
            cout<<res<<endl;
            continue;
        }
        else if(n<=m) cout<<1<<endl;
    }
    return 0;
}

2

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int n,m,t=1;
        cin>>n>>m;
        if(m==1){
            if(n==1)printf("1\n");
            else printf("-1\n");
        }
————————————————————————————————————————————————————————————————
 1. else{
                t=ceil(1.0*(n-m)/(m-1));//人麻了
                printf("%d\n",t*2+1);//咱就是说咱推了一下午花生都没搞出来这个式子
            }
            
 2.  else{
            time=(n-2)/(m-1)*2+1;
        }
        cout << time << endl;
 3.  else
        {
            LL ans=1;
            ans+=(n-2)/(m-1)*2;
            printf("%lld\n",ans);
        }
 4.  else
        {
            n -= m;
            count = (n + m - 2) / (m - 1);
            printf("%lld\n", count * 2 + 1);
        }
 5. else{
            int num1=n-1;
            int num2=m-1;
            if(num1%num2==0)cout<<((num1/num2-1)*2)+1<<endl;
            else cout<<((num1/num2)*2)+1<<endl;
        }
————————————————————————————————————————————————————————————————
    }
}

3

//康康的解法，感觉很迷惑55
int solve(){
     cin>>n>>m;
     if(n<=m)return 1;
     if(n>1&&m==1)return -1;
     int x=n-m,y=m+(x%(m-1)),z=(x/(m-1))*2;
     while(y>m){
         z+=2;
         y-=m-1;
       }
    return z+1;
}
int main(){
    cin>>t;
    while(t--)
    {cout<<solve()<<endl;}
    return 0;
}

F题
在这里插入图片描述

//一整个没想明白，来自我的撒比代码
#include<bits/stdc++.h>
using namespace std;
int t;
long long n,m;
const int N=1e5+10;
int a[N];
int main(){
    cin>>t;
    while(t--){
        cin>>n>>m;
        for(int i=0;i<n;i++)cin>>a[i];
        sort(a,a+n);
        if(n%2==1){
            if(a[n/2]<m){cout<<-1<<endl;continue;}
            else if(a[n/2]==m){cout<<1<<endl;continue;}
            else {cout<<n<<endl;continue;}
        }
        else{
            if(a[n/2-1]<m){cout<<-1<<endl;continue;}
            else if(a[n/2-1]==m){cout<<1<<endl;continue;}
            else if(a[n/2-1]>m){
                if(a[0]>=m)cout<<n<<endl;
                else cout<<1+n/2<<endl;
                continue;
            }
        }
    }
    return 0;
}

1

//永远能找到贼短的代码，但是看起来很有道理
#include<bits/stdc++.h>
using namespace std;
int main () {
    int n;
    cin >> n;
    while(n--) {
        int a,b,cot1 = 0,cot2 = 0;
        cin >> a >> b;
        int k[a];
        for(int i = 0; i < a; i++) {
         cin >> k[i];
            if(k[i] >= b)cot1++;
            else cot2++;
        }
        if(cot1 - cot2 <= 0)cout <<"-1"<<endl;
        else cout << cot1 - cot2 << endl;
        
    }
}

2

	int n;cin>>n;
    while(n--){
        ll a,b,cnt=0,t;
        cin>>a>>b;
        rep(i, 1, a){
            cin>>t;
            if(t<b)cnt++;
        }
        if(a-cnt*2>0)cout<<a-cnt*2<<endl;
        else cout<<-1<<endl;

3

void solve() {
    int n, m;
    cin >> n >> m;
    int l, r;
    l = r = 0;
    for(int i = 0; i < n; i ++) {
        int a ;
        cin >> a;
        if(a >= m) r ++;
        else l ++;
    }
    if(r <= l) cout << -1 << '\n';
    else cout << r - l << '\n';
}

4

 	int T;
    long ans,n,i;
    long long m,a[100001];
    cin>>T;
    while(T--)
    {
        scanf("%ld%lld",&n,&m);
        ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            if(a[i]>=m)
                ans++;
            else
                ans--;
        }
        if(ans<1)
            puts("-1");
        else
            printf("%ld\n",ans);

I题
在这里插入图片描述
1

//康子的代码
const int N = 200010;
const int mod = 1e9 + 7;

int tt = 1;

int pqow(ll b, ll n){//快速幂求逆元
    ll ret = 1;
    for(int i = n; i; i >>= 1, b = (b * b) % mod)
        if(i & 1) ret *= b, ret %= mod;
    return ret;
}

int solve(){
    ll n, m; cin >> n >> m;
    if(n == 1) return 0;
    ll fz = (pqow(2, n - 1) - 1) * m % mod;//数学期望求出来的式子求导得到
    ll fm = pqow(pqow(2, n - 1), mod - 2) % mod;
    return (fz * fm) % mod;
}

int main(){
	ios::sync_with_stdio(0);
	cin >> tt;
	while(tt--)cout << solve() << endl;
	return 0;
}

康康的教诲
在这里插入图片描述
2

ll quickpow(ll a, ll b) {
    ll ans = 1;
    ll res = a%mod;
    while (b) {
        if (b & 1)
            ans = ans * res%mod;
        b >>= 1;
        res = res * res%mod;
    }
    return ans%mod;
}
void solve() {
    int n,m; cin>>n>>m;
    ll a = (quickpow(2,n-1)-1+mod)%mod;
    a=a*m%mod;
    ll b = quickpow(2,n-1);
    ll ans = a*quickpow(b,mod-2)%mod;
    cout<<ans<<endl;
}

3

int qmi(int a,int b)
{
    int res=1;
    while(b)
    {
        if(b&1)res=res*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return res;
}
void solve()
{
        int n,m;
        int ans=1;
        cin>>n>>m;
        int k = qmi(2,n-1);
        int ni = qmi(k,mod-2);
        cout<<m*(1-ni+mod)%mod<<'\n';  
}

A题
1

#include<bits/stdc++.h>
using namespace std;
int n, f[100010][10], mod = 998244353;

int get(int x)
{
    if(x % 9 == 0)
        return 9;
    return x % 9;
}

int main()
{
    scanf("%d", &n);
    vector<int> a(n + 1);
    for(int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        a[i] = get(a[i]);
    }
    f[0][0] = 1;
    for(int i = 1; i <= n; i++)
        for(int j = 0; j <= 9; j++)
        {
            f[i][j] = (f[i][j] + f[i - 1][j]) % mod;
            int t = get(j + a[i]);
            f[i][t] = (f[i][t] + f[i - 1][j]) % mod;
        }
    for(int i = 1; i <= 9; i++)
        printf("%d ", f[n][i]);
    puts("");
    return 0;
}

2

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;

const double eps = 1e-6;
const double PI = acos(-1.0);
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 998244353;
int dx1[4] = {-1, 0, 1, 0}, dy1[4] = {0, 1, 0, -1};  //上右下左
int dx2[8] = {-1, -1, -1, 0, 1, 1, 1, 0}, dy2[8] = {-1, 0, 1, 1, 1, 0, -1, -1};  //从左上顺时针旋转
int dx3[8] = {-2, -1, 1, 2, 2, 1, -1, -2}, dy3[8] = {1, 2, 2, 1, -1, -2, -2, -1};  //日字，从右上顺时针

int T, n, m, k, ans;

int qmi(int a, int b, int p){int res = 1 % p;a %= p;while (b > 0){if(b & 1) res = (LL)res * a % p;a = (LL)a * a % p;b >>= 1;}return res;}
int doMod(int x){return (x % mod + mod) % mod;}
int lowbit(int x){return x & -x;}
LL gcd(LL a,LL b) {return b?gcd(b,a%b):a;}
void debug1(int x){cout << x << "====" << endl;}
void debug2(int x, int y){cout << x << "====" << y << endl;}
void debug3(int x, int y, int z){cout << x << "====" << y << "====" << z << endl;}
//void print1(){for (int i = 1; i <= n; i ++ ){printf("%d ", );}cout << endl;}
int w[N];
int f[N][10];
void print2(){for (int i = 1; i <= n; i ++ ){for (int j = 1; j <= 9; j ++ ) printf("%d ", f[i][j]);cout<<endl;}cout<<endl;}

int ff(int x)
{
    while (x >= 10)
    {
        int t = 0;
        while (x)
        {
            t += x % 10;
            x /= 10;
        }
        x = t;
    }
    
    return x;
}

int main()
{
// 	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
//     exit(0);
    cin >> n;
    for (int i = 1; i <= n; i ++ ) cin >> w[i];
    f[0][0] = 1;
    for (int i = 1; i <= n; i ++ )
    {
        for (int j = 0; j <= 9; j ++ )
        {
            f[i][ff(j + w[i])] = (f[i][ff(j + w[i])] + f[i - 1][j]) % mod;
            f[i][j] = (f[i][j] + f[i - 1][j]) % mod;
        }
    }
    for (int i = 1; i < 10; i ++ )
    {
        cout << f[n][i] << ' ';
    }
    return 0;
}

3

using namespace std;
const int mod = 998244353;
const int maxn = 1e5 + 5;
int a[maxn];
int dp[maxn][10];
int main()
{
	int n;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		a[i] %= 9;
	}
	dp[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= 9; j++) {
			dp[i][j] =( dp[i][j]+dp[i - 1][j])%mod;
			dp[i][(j+a[i])%9]= (dp[i][(j + a[i]) % 9] + dp[i - 1][j]) % mod;
		}
	}
	for (int i = 1; i <= 8; i++) cout << dp[n][i] << " ";
	cout << dp[n][0] - 1 << endl;
	return 0;
}

4

#include <cstdio>
#include <iostream>
using namespace std;
const int mod = 998244353;
const int maxn = 1e5 + 5;
int a[maxn];
int dp[maxn][10];
int main()
{
	int n;cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		a[i] %= 9;
	}
	dp[0][0] = 1;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j <= 9; j++) {
			dp[i][j] =( dp[i][j]+dp[i - 1][j])%mod;
			dp[i][(j+a[i])%9]= (dp[i][(j + a[i]) % 9] + dp[i - 1][j]) % mod;
		}
	}
	for (int i = 1; i <= 8; i++) cout << dp[n][i] << " ";
	cout << dp[n][0] - 1 << endl;
	return 0;
}

J题
在这里插入图片描述

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)
#define pb push_back
#define SZ(v) ((int)v.size())
#define fs first
#define sc second
typedef long long ll;
typedef double db;
typedef pair<int,int> pii;
int T,A,B,n;
int va[10010],vb[10010];
bool cmp(int xx,int yy){
    return xx>yy;
}
int suma[10010],sumb[10010];
int main(){
    cin>>T;
    while(T--){
        cin>>A>>B>>n;
        rep(i,1,A)    scanf("%d",&va[i]);
        rep(i,1,B)    scanf("%d",&vb[i]);
        int mxb=min(n/2,B);
        if(A+mxb<n){
            puts("-1");
            continue;
        }
        sort(va+1,va+1+A,cmp);
        sort(vb+1,vb+1+B,cmp);
        rep(i,1,A)    suma[i]=suma[i-1]+va[i];
        rep(i,1,B)    sumb[i]=sumb[i-1]+vb[i];
        int ans=-1;
        rep(ia,0,A){
            int ib=n-ia;
            if(ib>mxb||ib<0)    continue;
            ans=max(ans,suma[ia]+sumb[ib]);
        }
        assert(ans!=-1);
        printf("%d\n",ans);
    }
    return 0;
}

2

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int gg[10001],hh[10001];
int t,i,a,b,sum = 0,n,j;
bool cmp(int x,int y)
{
	return x>y;
}
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		cin>>a>>b>>n;
		for(i=1;i<=a;i++)
		{
			scanf("%d",&gg[i]);
		}
		for(i=1;i<=b;i++)
		{
			scanf("%d",&hh[i]);
		}
		if(2*a<n||a+b<n)//把所有乖小孩都选了还是不够
		{
			printf("-1\n");
			continue;	
		}
		int sum1[10001] = {0},sum2[10001] = {0};
		sort(gg+1,gg+1+a,cmp);//将安静小孩从大到小排序
		sort(hh+1,hh+1+b,cmp);//将闹腾小孩幸福指数从大到小排序
		sum = 0;
		i = 1;
		j = 1;
		for(i=1;i<=a;i++)//正着将每多一个小孩幸福指数增加多少给找出来存入数组 
		{
			sum1[i]+=sum1[i-1]+gg[i];
		}
		for(i=1;i<=b;i++)
		{
			sum2[i]+=sum2[i-1]+hh[i];
		}
		sum = -1;
		for(i=0;i<=a;i++)//从0开始，因为有可能一个不选
		{
			if((n-i)>b||n<i||(n-i)>n/2)
			{
				continue;	
			}
			sum = max(sum,sum1[i]+sum2[n-i]);	
		}
		printf("%d\n",sum);  
	}
	return 0;
}
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