rmq_st实现
const int MAXN = 12;
int num[MAXN] = {1,5,3,2,7,9,3,6,7,0,6,8};
int sparseTable[MAXN][MAXN];
inline int Min(int a,int b)
{if(a<b)return a;elsereturn b;
}
void st(int n)//rmq的sparse table 的預處理
{memset(sparseTable,0,sizeof(sparseTable));for(int i=0;i<MAXN;++i)sparseTable[i][0] = num[i];int k = log(MAXN*1.0)/log(2.0);for (int i=1;i<=k;++i){for (int j =0;j+(1<<i)<=MAXN;++j){sparseTable[j][i] = Min(sparseTable[j][i-1], sparseTable[j+(1<<(i-1))][i-1]);}}
}int rmq(int beg,int end)
{int k = log( (double)(end - beg+1))/log(2.0);return Min(sparseTable[beg][k],sparseTable[beg+(1<<(k-1))][k]);//貌似有點問題:Min(sparseTable[beg][k],sparseTable[end -(1<<k)+1)][k]}
int main()
{st(MAXN);cout<<rmq(1,11)<<endl;
}
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