第28章：最大连续乘积子串

-2.5 4 0 3 0.5 8 -1Ans:(3,0.5,8)

template<typenameComparable>

Comparable maxprod(constvector<Comparable>& v)

{

int i;

Comparable maxProduct =1;

Comparable minProduct =1;

Comparable maxCurrent = 1;

Comparable minCurrent =1;

for(inti=0;i<v.size();i++)

{

maxCurrent *= v[i];

minCurrent *= v[i];

if(maxCurrent >maxProduct)

maxProduct =maxCurrent;

if(minCurrent >maxProduct)

maxProduct =minCurrent;

if(maxCurrent <minProduct)

minProduct =maxCurrent;

if(minCurrent <minProduct)

minProduce =minCurrent;

if(minCurrent >maxCurrent)

Swap(maxCurrent,minCurrent);

if(maxCurrent < 1)

maxCurrent = 1;

return maxProduct;

}

}

*直接用动态规划求解

状态转移方程：max表示以a结尾的最大连续子串

Max = max{a,Max[i-1]*a,Min[i-1]*a};

Min =min{a,Max[i-1]*a,Min[i-1]*a};

C++代码：

double func(double* a ,constint n)

{

double* maxA = newdouble[n];

double* minA = newdouble[n];

msxA[0] = min[A] = a[0];

double value = maxA[0];

for(int i=1;i<n;i++)

{

maxA[i] =max(max(a[i],maxA[i-1]*a[i]),minA[i-1]*a[i]);

minA[i] =min(,min(a[i],maxA[i-1]*a[i]),minA[i-1]*a[i]);

vlaue =max(value,maxA[i]);

}

delete [] maxA;

delete [] minA;

return value;

}