exercise3

第三次作业

1、两名参与者交替从一堆石子中取出若干数目,其个数由参与者自已决定.但是要求参与者每次至少取出一个,至多取出一半,然后另一名参与者继续.拿到最后一个石子的参与者将输掉该游戏.

输入：输入开始时石子的个数

输出：如果先取者输，则输出“lose”，否则输出“win”

#include

#include

using namespace std;

int main(){

int number;

cin>>number;

double i=1;

double k=pow(2.0,i)-1;

while(k

++i;

k=pow(2.0,i)-1;

}

if(k==number)

cout<<"lose";

if(k>number)

cout<<"win";

return 0;

}

2、Given some Chinese Coins (three kinds-- 1, 2, 5), and their number is num_1, num_2 and num_5 respectively, please output the minimum value

that you cannot pay with given coins.

Input

Input contains multiple test cases. Each test case contains 3 positive integers num_1, num_2 and num_5 (0<=num_i<=1000). A test case containing 0 0 0 terminates the input and this test case is not to be processed.

Output

Output the minimum positive value that one cannot pay with given coins, one line for one case.

#include

#include

#include

using namespace std;

const int MAX = 1000;

const int type[3] = {1,2,5};

bool ans[MAX]={0},tmp[MAX]={0};

int cnt[3];

void Search(){

int i,j,k;

for(i=0;i<=cnt[0];++i){

ans[i]=true;

}

for(i=1;i<3;++i){

for(j=0;j

for(k=0;k<=cnt[i] && j+k*type[i]<=MAX;++k){

if(tmp[j+k*type[i]] || ans[j])

tmp[j+k*type[i]]=true;

}

}

for(j=1;j<=MAX;++j){

ans[j]=tmp[j];

tmp[j]=false;

}

}

}

int main(){

int res=0;

scanf("%d %d %d",&cnt[0],&cnt[1],&cnt[2]);

if(cnt[0]==0 && cnt[1]==0 && cnt[2]==0){

printf("0");

return 0;

}

Search();

for(int i=0;i<=MAX;++i){

if(!ans[i]){

res=i;

break;

}

}

printf("%d",res);

return 0;

}

3、写出完整的二分查找程序，要求：

（1）先将给定的数据利用快速排序进行排序（采用复杂度为nlg(n)的算法）。

（2）利用二分查找法判断给定的数据是否在数据集中。

#include

#include

using namespace std;

void quicksort(int a[],int left,int right){

if(left>=right)

return;

else{

int first=left;

int last=right;

int key=a[first];

while(first

{

while(first=key)

--last;

a[first]=a[last];

while(first

++first;

a[last]=a[first];

}

a[first]=key;

quicksort(a,left,first-1);

quicksort(a,first+1,right);

}

}

int search(int num,int a[],int size){

int left=0;

int right=size-1;

int mid=(left+right)/2;

while(left<=right){

if(a[mid]==num)

return mid;

else if(a[mid]>num)

right=mid-1;

else

left=mid+1;

mid=(left+right)/2;

}

return 0;

}

int main(){

int s[]={2,1,5,87,43,23,9};

quicksort(s,0,6);

int num=23;

if(search(num,s,7))

printf("给定的数据在数据集中!");

else

printf("给定的数据不在数据集中!");

return 0;

}

4、

Problem Description

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings.

Input

The first line of the input file contains a single integer t (1 <= t <= 10), the number of test cases, followed by the input data for each test case. The first line of each test case contains a single integer n (1 <= n <= 100), the number of given strings, followed by n lines, each representing one string of minimum length 1 and maximum length 100. There is no extra white space before and after a string.

Output

There should be one line per test case containing the length of the largest string found.