第十四届蓝桥杯三月真题刷题训练——第 3 天（3.6）

2023-03-05 21:32:40 79 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

SRAM compiler中Power Gating与Retention的实战配置指南本文深入解析了在SRAM compiler中配置Power Gating（电源门控）与Retention（数据保持）功能的实战指南。针对低功耗芯片设计，文章详细阐述了如何根据应用场景（如移动AP、IoT设备）权衡开关类型、尺寸、唤醒时序以及保持电压，以实现静态功耗优化与数据可靠性的最佳平衡，并强调了后端实现与签核检查的关键要点。阅读详情

门牌制作 - 蓝桥云课 (lanqiao.cn)

货物摆放 - 蓝桥云课 (lanqiao.cn)

跳跃 - 蓝桥云课 (lanqiao.cn)

重新排序 - 蓝桥云课 (lanqiao.cn)

门牌制作 - 蓝桥云课 (lanqiao.cn)

#include <bits/stdc++.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <set>
#define x first
#define y second
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef pair<int,int> PII;
typedef pair<char,int> PCI;
typedef long long LL;
//typedef __int128 i128;
typedef unsigned long long ULL;
const int N=4e5+10,INF = 0x3f3f3f3f ;
const double eps = 1e-7;

LL res= 0;

int cnt(int x)
{
	int c=0,t;
	while(x) 
	{
		t = x % 10;x/=10;
		if(t == 2 ) c ++ ;
	}
	return c;
}

int main()
{
	for(int i=1;i<=2020;i ++ )
	{
		res += cnt(i);
	}
	cout << res <<endl;
	return 0;
}

货物摆放 - 蓝桥云课 (lanqiao.cn)

#include <iostream>
#include <bits/stdc++.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>
#define x first
#define y second
#define endl '\n'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<LL,LL> PLL;
//typedef __int128 i128;
typedef unsigned long long ULL;
const int N= 20000 +  10,M = 55 ;
const int INF = 0x3f3f3f3f ,mod = 1e9 + 7 ,P = 131;
const double eps = 1e-7;

LL n = 2021041820210418;
LL cnt,p[N];


void solve()
{
	for(LL i =1;i <= n/i;i ++ )
	{
		if(n % i == 0)
		{
			p[cnt ++ ] = i;
			if(n/i != i )
			{
				p[cnt ++ ] = n/i;
			}
		}
	}
	LL res=0;
	for(LL i=0;i<cnt ;i++ )
	for(LL j=0;j<cnt;j ++ )
	for(LL k= 0;k< cnt ;k ++ )
	{
		res += (p[i] * p[j] * p[k] == n );
	}
	
	cout << res <<endl;
	
		
}

signed main()
{
//	freopen("1.txt","r",stdin);
    
    
    int T=1;
//    cin>>T;
	int now = 1;
    while(T -- )
    {
//    	cout<<"Case "; 
//    	cout<< now ++ <<": ";
        solve();
    }


    return 0;
}

跳跃 - 蓝桥云课 (lanqiao.cn)

#include <iostream>
#include <bits/stdc++.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>
#define x first
#define y second
#define endl '\n'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<LL,LL> PLL;
//typedef __int128 i128;
typedef unsigned long long ULL;
const int N= 100+  10,M = 55 ;
const int INF = 0x3f3f3f3f ,mod = 1e9 + 7 ,P = 131;
const double eps = 1e-7;

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



void solve()
{
	cin >> n >> m ;
	
	for(int i=3;i<=n + 2 ;i ++ )
	for(int j=3;j<=m +2; j ++ )
	cin >> w[i][j];
	ms(f,-0x3f);
	
	f[3][3] = w[3][3];
	for(int i=3;i<=n + 2 ;i ++ )
	for(int j=3;j<=m +2; j ++ )
	{
		if(i == 3 && j == 3 ) continue;
		int &t = f[i][j];
		
		for(int x=1;x<=3 && i -x >= 3 ;x ++ )
		{
			t =max(t,f[i-x][j]);
		}
		
		for(int y=1;y <= 3 && j - y >= 3 ; y ++ )
		{
			t = max(t,f[i][j-y]);
		}
		t += w[i][j];
	}
	cout << f[n + 2 ][m + 2 ]  <<endl; 
	
		
}

signed main()
{
//	freopen("1.txt","r",stdin);
    
    
    int T=1;
//    cin>>T;
	int now = 1;
    while(T -- )
    {
//    	cout<<"Case "; 
//    	cout<< now ++ <<": ";
        solve();
    }


    return 0;
}

重新排序 - 蓝桥云课 (lanqiao.cn)

#include <iostream>
#include <bits/stdc++.h>
#include <cstring>
#include <algorithm>
#include <vector>
#include <set>
#include <queue>
#define x first
#define y second
#define endl '\n'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;

typedef long long LL;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<LL,LL> PLL;
//typedef __int128 i128;
typedef unsigned long long ULL;
const int N= 100+  10,M = 55 ;
const int INF = 0x3f3f3f3f ,mod = 1e9 + 7 ,P = 131;
const double eps = 1e-7;


LL n,m,a[N],s[N],b[N],c[N];
LL s1,s2,res;

void solve()
{
	cin >> n;

	for(int i=1;i <= n; i++ )cin >> a[i],s[i] = s[i-1] + a[i];
	
	cin >> m;
	while(m--)
	{
		int l,r;cin >> l >> r;
		b[l] ++ ,b[r + 1] --;
		s1 += s[r] - s[l-1];
	}
	
	for(int i=1;i<= n;i ++ ) c[i] = c[i-1] + b[i];
	
	sort(c + 1, c + n + 1);// 将每个数的出现次数排序 
	sort(a + 1, a + n + 1);// 将原来每个数大小排序 
	
	for(int i=1;i<=n;i ++ )
	{
		s2 += c[i] * a[i]; 
		// 重新分配每个数出现的次数 ，数值小的出现次数少，数值大的出现次数多 
	}
	
	
	cout << s2 - s1 << endl;
}

signed main()
{
//	freopen("1.txt","r",stdin);
    
    
    int T=1;
//    cin>>T;
	int now = 1;
    while(T -- )
    {
//    	cout<<"Case "; 
//    	cout<< now ++ <<": ";
        solve();
    }


    return 0;
}

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