计算机找工作笔试题集锦(详解)_算法问题

1.

题目射击运动员10发打中90环有多少种可能，请编写程序计算出来，并打印出结果，0环和10环均有效

答案递归算法：

#include

#include

int f(int n, int m)

{ //n次射中m环

int fn=0,i;

if(m<0||m>10*n) return 0;

if(n==1) return 1;

for(i=0;i<=10;i++) //本次射中的为i环，剩下的n－1次要射m－i环

fn+=f(n-1,m-i);

return fn;

}

int main()

{

int n,m;

printf("Please enter n and m:");

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

printf("\n%d发打中%d环有%d种可能.\n",n,m,f(n,m));

system("PAUSE");

return 0;

}

非递归算法：

算法分析：

由于中90环和中10环的概率是一样的 (可用数学归纳法证明)

所以在此只需穷举和10的个数即可。

count = 0;

for(a=0;a<=10;a++)

for(b=0;b<=10-a;b++)

for(c=0;c<=10-a-b;c++)

for(d=0;d<=10-a-b-c;d++)

for(e=0;e<=10-a-b-c-d;e++)

for(f=0;f<=10-a-b-c-d-e;f++)

for(g=0;g<=10-a-b-c-d-e-f;g++)

for(h=0;h<=10-a-b-c-d-e-f-g;h++)

for(i=0;h<=10-a-b-c-d-e-f-g-h;i++)

for(j=0;h<=10-a-b-c-d-e-f-g-h-i;j++)

{

count++;

}

printf("%d\n",count);

int GetFifoBuffer(int size, void *data);

int FifoBuffer::GetQNodeBuffer(int size, FIFOQNode *node, void * data);

int GetSize();

private:

int m_size;

FIFOQNode* m_QHead;

FIFOQNode* m_QTail;

};

FifoBuffer::FifoBuffer()

{

m_size = 0;

m_QHead = m_QTail = NULL;

}

FifoBuffer::~FifoBuffer()

{

while(m_QHead != NULL)

{

FIFOQNode *p = m_QHead;

m_QHead = m_QHead->next;

delete [] p->data;

delete p;

}

}

bool FifoBuffer::PutFifoBuffer(int size, void *data)

{

if(size <= 0)

return false;

if(m_size + size > MAXBUFFER)

return false;

m_size += size;

char *newData = new char[size];

memcpy(newData, (void*)data, size);

FIFOQNode* pNewNode = new FIFOQNode();

pNewNode->data = newData;

pNewNode->size = size;

pNewNode->next = NULL;

if(m_QHead == NULL)

m_QHead = m_QTail = pNewNode;

else

{

m_QTail->next = pNewNode;

m_QTail = pNewNode;

}

return true;

}

int FifoBuffer::GetQNodeBuffer(int size, FIFOQNode *node, void * data) {

if(size <= 0 || size > node->size || node == NULL)

return 0;

memcpy(data, node->data, size);

FIFOQNode *p = node;

if(size < node->size)

{

char *newData = new char[node->size - size];

memcpy((void*)newData, &node->data[size], node->size - size);

FIFOQNode* pNewNode = new FIFOQNode();

pNewNode->data = newData;

pNewNode->size = node->size - size;

pNewNode->next = node->next;

node = pNewNode;

}

else

{

node = node->next;

}

delete [] p->data;

delete p;

return size;

}

int FifoBuffer::GetFifoBuffer(int size, void *data)

{

if(size <= 0)

return 0;

if(size > m_size)

size = m_size;

FIFOQNode *curNode = m_QHead;

int remainSize = size;

while(curNode != NULL && remainSize > curNode->size)

{

int cpSz = GetQNodeBuffer(curNode->size, curNode,

(void*)&((char*)data)[size - remainSize]);

remainSize -= cpSz;

}

if(remainSize > 0)

GetQNodeBuffer(remainSize, curNode, (void*)&((char*)data)[size - remainSize]);

m_size -= size;

void ClearPQ(void);

//test op

int PQEmpty(void) const;

int PQFull(void) const;

int PQLength(void) const;

};

//PQInsert

template

void PQueue::PQInsert(const DataType &item)

{

//if full, then exit

if(count == MaxPQSize)

{

perror("Priority queue overflow!");

exit(1);

}

// put the element at the end of the queue

pqlist[count ++] = item;

}

//PQDelete

template

DataType PQueue::PQDelete()

{

DataType min;

int i, minindex = 0;

if(count > 0)

{

//find the index of the min element

min = pqlist[0];

for(i = 1; i < count; i ++)

{

if(pqlist[i] < min)

{

min = pqlist[i];

minindex = i;

}

}

//put the element in the end to the index of the min

pqlist[minindex] = pqlist[count - 1];

count --;

}

//if is empty

else

{

}

int LevenshteinDistance(char *str1, char *str2)

{

// d is a table with lenStr1+1 rows and lenStr2+1 columns int len1 = strlen(str1);

int len2 = strlen(str2);

//以下申请一个 (len1+1)*(len2+1)的二维数组

int **d = new int*[len1 + 1];

for(int k = 0; k < len1 + 1; k ++)

d[k] = new int[len2 + 1];

// i and j are used to iterate over str1 and str2

int i, j, cost;

for(i = 0; i < len1 + 1; i ++)

d[i][0] = i;

for(j = 0; j < len2 + 1; j ++)

d[0][j] = j;

for(i = 1; i <= len1; i ++)

for(j = 1; j <= len2; j++)

{

cost = (str1[i] == str2[j] ? 0 : 1);

d[i][j] = my_min(d[i-1][j ] + 1, // deletion

d[i][j-1] + 1, // insertion

d[i-1][j-1] + cost); // substitution }

cout << "result:" << endl; //输出矩阵

for(i = 0; i <= len1; i ++)

{

for(j = 0; j <= len2; j++)

{

cout << d[i][j] << " ";

}

cout << endl;

}

return d[len1][len2]; //返回值为编辑距离

}

int main(int argc, char* argv[])

{

int* p = new int(5);

char *str1 = "z11zjw";

char *str2 = "z12zjw";

LevenshteinDistance(str1, str2);

}