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-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;
}
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原文地址:http://www.cnblogs.com/lsx1993/p/4841552.html