/*可以看出，对于一段区间[L,R]如果统计了答案若a[L]
     
      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
      
       #define N 1000007#define ll long longusing 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<
       
        <
       
      
     

 

 

/*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
     
      #define mod 998244353#define N 10005using 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<
     

 

#include
      
       #define N 30#define ll long long#define mod 1000000007#define opt 99999using 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]
      

#include 
      
       #include 
       
        #include 
        
         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
         
           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);}
         
        
       
      

暂时放弃  先做完互不侵犯king再来写这道题