10.11NOIP模拟题(3)

最新推荐文章于 2024-11-08 16:11:58 发布

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最新推荐文章于 2024-11-08 16:11:58 发布

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原文链接：http://www.cnblogs.com/L-Memory/p/9775325.html

本文探讨了两种算法问题：一是求解区间内的最大值问题，通过双指针技巧实现；二是统计一组数中二进制位1的分布，优化平方和计算，涉及位运算和数据统计。

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/*
可以看出，对于一段区间[L,R]如果统计了答案
若a[L]<a[R],那么当右端点往左移时答案不会更优，a[R]>a[L]同理
所以两个指针分别从头尾往中间扫那边小移哪边即可。
*/
#include<bits/stdc++.h>

#define N 1000007
#define ll long long

using namespace std;
ll a[N];
ll n,m,ans,cnt;

int main()
{
    freopen("w.in","r",stdin);
    freopen("w.out","w",stdout);
    scanf("%lld",&n);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    int L=1,R=n;
    while(L<=R)
    {
        ll res=(R-L+1)*min(a[L],a[R]);
        ans=max(ans,res);
        if(a[L]>=a[R]) R--;
        else L++;
    }
    cout<<ans<<endl;
    return 0;
}

/*
发现算来算去所有数的二进制1的个数和位置不会变。
两个数and 或 or只是把他们的二进制位的1聚拢到一个数上。
若两个数为a，b，假设他们进行一次操作后变为(a+x),(b-x),则a>b
那么(a+x)^2+(b-x)^2=a^2+b^2+2x(x+a-b)>a^2+b^2
说明这种“聚拢”操作后平方和会变大
所以把所有数的二进制位1统计下来后尽可能的对分给一个数即可。
*/
#include<bits/stdc++.h>

#define mod 998244353
#define N 10005

using namespace std;
int n,x;
int cnt[N];

int main()
{
    freopen("s.in","r",stdin);
    freopen("s.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        for (int j=0;j<32;j++) if (x & (1<<j)) cnt[j]++;
    }
    unsigned long long ans=0;
    for(int i=0;i<n;i++)
    {
        x=0;
        for(int j=0;j<32;j++) if (cnt[j]) x |=(1<<j),cnt[j]--;
        ans+=(unsigned long long)(x)*x%mod;;
    }
    cout<<ans%mod;
    return 0;
}

#include<bits/stdc++.h>

#define N 30
#define ll long long
#define mod 1000000007
#define opt 99999

using namespace std;
int n,m,k,lim;
ll ans;
int dx[4]={0,0,1,-1};
int dy[4]={1,-1,0,0};
int vis[N][N],f[N][N];

inline ll read()
{
    int x=0,f=1;char c=getchar();
    while(c>'9'||c<'0'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

bool can(int x,int y)
{
    if(vis[x][y]==-1) return false;
    for(int i=max(1,y-2);i<=min(y+2,m);i++)
      if(vis[x][i]>=opt) return false;
    for(int i=max(1,x-2);i<=min(n,x+2);i++)
      if(vis[i][y]>=opt) return false;
    if(vis[x-1][y-1]>=opt || vis[x-1][y+1]>=opt) return false;
    if(vis[x+1][y+1]>=opt || vis[x+1][y-1]>=opt) return false;
    return true;
}

void update(int x,int y)
{
    vis[x][y]=opt;
    for(int i=max(1,y-2);i<=min(y+2,m);i++)
      if(vis[x][i]!=-1)vis[x][i]++;
    for(int i=max(1,x-2);i<=min(n,x+2);i++)
      if(vis[i][y]!=-1)vis[i][y]++;
    if(x-1 && y-1 && vis[x-1][y-1]!=-1) vis[x-1][y-1]++;
    if(x-1 && y+1<=m && vis[x-1][y+1]!=-1) vis[x-1][y+1]++;
    if(x+1<=n && y-1 && vis[x+1][y-1]!=-1) vis[x+1][y-1]++;
    if(x+1<=n && y+1<=m && vis[x+1][y+1]!=-1) vis[x+1][y+1]++;
}

void _update(int x,int y)
{
    vis[x][y]=0;
    for(int i=max(1,y-2);i<=min(y+2,m);i++)
      if(vis[x][i]>0)vis[x][i]--;
    for(int i=max(1,x-2);i<=min(n,x+2);i++)
      if(vis[i][y]>0)vis[i][y]--;
    if(vis[x-1][y-1]>0)vis[x-1][y-1]--;
    if(vis[x-1][y+1]>0)vis[x-1][y+1]--;
    if(vis[x+1][y-1]>0)vis[x+1][y-1]--;
    if(vis[x+1][y+1]>0)vis[x+1][y+1]--;    
    f[x][y]=0;
}

void dfs(int x,int y,int L)
{
    if(L==0)
    {
        ans++;ans%=mod;
        return;
    }
    for(int xx=x;xx<=n;xx++) for(int yy=1;yy<=m;yy++)
    {
        if(f[xx][yy]) continue;
        if(vis[xx][yy]<opt && can(xx,yy) && !f[xx][yy])
        {
            update(xx,yy);
            f[xx][yy]=1;
            dfs(xx,yy,L-1);
            _update(xx,yy);//dfs(xx,yy,L);
        }
    }
}

int main()
{
    freopen("l.in","r",stdin);
    freopen("l.out","w",stdout);
    int x,y;ans=0;
    n=read();m=read();k=read();
    for(int i=1;i<=n;i++)
    {
        x=read();y=read();
        vis[x][y]=-1;
    }
    lim=n*m-k;
    while(lim--)
    {
        memset(f,0,sizeof f);
        f[1][1]=1;
        dfs(1,1,lim);
    }
    ans%=mod;cout<<ans<<endl;
    return 0;
}

没调出来的dfs

#include <cstdio>
#include <cstring>
#include <vector>

const int MAXN = 15;
const int MOD = 1e9 + 7;

int m;

inline unsigned int getBit(int i) {
    return 1u << i;
}

inline bool isValidLine(unsigned int s) {
    return !((s & (s << 1)) || (s & (s << 2)));
}

inline bool isValidTwoLines(unsigned int a, unsigned int b) {
    return !((a & (b << 1)) || (a & (b >> 1)) || (a & b));  
}

inline bool isValidThreeLines(unsigned int a, unsigned int b, unsigned int c) {
    return !(a & c) && isValidTwoLines(a, b)/* && isValidTwoLines(b, c)*/;
}

int main() {
    freopen("l.in", "r", stdin);
    freopen("l.out", "w", stdout);

    int n, k;
    scanf("%d %d %d", &n, &m, &k);

    unsigned int ban[n + 1];
    memset(ban, 0, sizeof(ban));
    while (k--) {
        int i, j;
        scanf("%d %d", &i, &j);
        ban[i] |= 1 << (j - 1);
    }

    std::vector<unsigned int> validStates;

    unsigned int maxS = 1 << m;
    for (int i = 0; i < maxS; i++) {
        if (isValidLine(i)) {
            validStates.push_back(i);
        }
    }

    // f[i][state of line i - 1][state of line i] = count
    long long f[n + 1][validStates.size()][validStates.size()];

    memset(f, 0, sizeof(f));
    for (int a = 0; a < (int)validStates.size(); a++) f[0][0][a] = 1;
    
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int c = 0; c < (int)validStates.size(); c++) {
            if (ban[i] & validStates[c]) continue;
            for (int b = 0; b < (i <= 1 ? 1 : (int)validStates.size()); b++) {
                if (!isValidTwoLines(validStates[b], validStates[c])) continue;
                if (ban[i - 1] & validStates[b]) continue;

                for (int a = 0; a < (i <= 2 ? 1 : (int)validStates.size()); a++) {
                    if (!isValidThreeLines(validStates[a], validStates[b], validStates[c])) continue;
                    if (a && (ban[i - 2] & validStates[a])) continue;

                    (f[i][b][c] += f[i - 1][a][b]) %= MOD;
                }

                if (i == n) (ans += f[i][b][c]) %= MOD;
            }
        }
    }

    printf("%lld\n", ans);
}