南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解

时间:2021-08-11 23:28:07

A,C,I签到题,只搞了8题,还一题是神仙做的,我不会

 

链接:https://www.nowcoder.com/acm/contest/122/B
来源:牛客网

取石子
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

现在有两堆石子,两个人轮流从中取石子,且每个人每一次只能取1、3或9个石子,取到最后一个石子的人win。
假设先手后手都会选择最好的方式来取石子,请您判断先后手的输赢情况。

输入描述:

多组输入
每组一行,一行包括两个正整数n1和n2(1<=n1<=100,1<=n2<=100),代表了两堆石子的数目

输出描述:

如果先手能赢,输出"win";否则就输出"lose"。
示例1

输入

1 1 
1 2

输出

lose
win

分析:这题挺有搞笑,两堆加起来是奇数就win,否则lose。

南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解
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 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FFI(a, b) for(int i = a; i < b; i++)
 76 #define FFJ(a, b) for(int j = a; j < b; j++)
 77 #define RR(a, b) for(int i = a; i > b; i++)
 78 #define ME(a, b) memset(a, b, sizeof(a))
 79 #define SC(x) scanf("%d", &x)
 80 #define PR(x) printf("%d\n", x)
 81 #define INF 0x3f3f3f3f
 82 #define MAX 1001
 83 #define MOD 1000000007
 84 #define E 2.71828182845
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 const double PI = acos(-1.0);
 89 typedef pair<int, int> Author;
 90 vector<pair<string, int> > VP; 
 91 
 92 int main(void){
 93     #ifdef LOCAL
 94         freopen("in.txt", "r", stdin);
 95         freopen("out.txt", "w", stdout);
 96     #endif
 97     ios::sync_with_stdio(false); cin.tie(0);
 98     int a, b;
 99     
100     while(cin>>a>>b){
101         if((a + b) % 2) cout<<"win"<<endl;
102         else cout<<"lose"<<endl;
103     }
104     return EXIT_SUCCESS;
105 }
View Code

链接:https://www.nowcoder.com/acm/contest/122/D
来源:牛客网

小C的记事本
时间限制:C/C++ 2秒,其他语言4秒
空间限制:C/C++ 131072K,其他语言262144K
64bit IO Format: %lld

题目描述

小C最近学会了java小程序的开发,他很开心,于是想做一个简单的记事本程序练练手。

他希望他的记事本包含以下功能:

1、append(str),向记事本插入字符串 str(英文字符)

2、delete(k),删除记事本最后k个字符(保证不为空串)

3、print(k),输出记事本第k个字符(保证不为空串)

4、undo(),撤销最近的1(或者)操作,使记事本回到1(或者2)操作之前的状态

可怜的小C琢磨了半天还是做不来,聪明的你能解决小C的问题吗?

输入描述:

多组输入

第一行输入一个整数q,代表操作总数

以下q行每行描述了一个操作,每行以一个整数t开始(1 <= t <= 4)。

t表示上述问题陈述中定义的操作类型。 如果操作需要参数,则后跟空格分隔的参数。

题目保证所有操作均合法

1 <= q <= 10^6 
1 <= k <= |记事本内容长度| 
每个测试数据中str的总长度 <= 10^6

请使用 ios::sync_with_stdio(false); 对读写进行加速

输出描述:

所有操作类型3必须输出第k个字符,每行以换行符结束。
示例1

输入

8
1 ab
3 2
2 2
1 cd
3 1
4
4
3 1

输出

b
c
a

说明

**样例解释**

假设记事本用字符串S表示

1、插入ab,S="ab"

2、输出第2个字符,是b

3、删除最后2个字符,S=""

4、插入cd, S="cd"

5、输出第1个字符,是c

6、撤销,此时S=""

7、撤销,此时S="ab"

8、输出第1个字符,是a

分析:这题很实用啊,现在都很喜欢回到上一步,和武大校赛那题有点像,不过这里是用stack解决的,因为可以回到之前所有的步骤,

和电脑上ctrl+z有点像,每次保存/删除输入字符串,str += str1,对于1,2进行一步操作就压栈,最后s.pop(),即可

南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解
  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
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 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FFI(a, b) for(int i = a; i < b; i++)
 76 #define FFJ(a, b) for(int j = a; j < b; j++)
 77 #define RR(a, b) for(int i = a; i > b; i++)
 78 #define ME(a, b) memset(a, b, sizeof(a))
 79 #define SC(x) scanf("%d", &x)
 80 #define PR(x) printf("%d\n", x)
 81 #define INF 0x3f3f3f3f
 82 #define MAX 1001
 83 #define MOD 1000000007
 84 #define E 2.71828182845
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 const double PI = acos(-1.0);
 89 typedef pair<int, int> Author;
 90 vector<pair<string, int> > VP; 
 91 
 92 int main(void){
 93     #ifdef LOCAL
 94         freopen("in.txt", "r", stdin);
 95         freopen("out.txt", "w", stdout);
 96     #endif
 97     ios::sync_with_stdio(false); cin.tie(0);
 98     int n, command, temp;string str, str1;
 99     
100     while(cin>>n){
101         stack<string> s; str = "";s.push(str);
102         while(n--){
103             cin>>command;
104             if(command == 1){
105                 cin>>str1;
106                 str += str1;
107                 s.push(str);
108             }
109             else if(command == 2){
110                 cin>>temp;
111                 str.erase(str.size() - temp, temp); 
112                 s.push(str);
113             }
114             else if(command == 3){
115                 cin>>temp;
116                 cout<<str[temp - 1]<<endl;
117             }
118             else{
119                 s.pop();
120                 str = s.top();
121             }
122          }
123     }
124     
125     return EXIT_SUCCESS;
126 }
View Code

 

 

链接:https://www.nowcoder.com/acm/contest/122/F
来源:牛客网

阿汤的疑惑
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

阿汤同学最近刚学数论,他发现数论实在是太有趣了,于是他想让你也感受一下数论的乐趣。现在他给你一个正整数 N 和一个正整数 M,要求你用 N 对 M 进行取余操作,即 N % M,记余数为 S。
但是他发现这样好像并不能让你感受到数论的乐趣,于是他想让你在N 对 M 取余操作的基础上再求出这个余数 S 能分解出多少个不同质因数。

质因数:质因数在数论里是指能整除给定正整数的质数,质数就是只能整除 1 和本身的数,定义 2 是最小的质数。

输入描述:

从标准输入读入数据。
输入包含多组数据,第一行一个整数 T 代表数据组数。接下来依
次描述每组数据,对于每组数据:
第一行输入正整数 N,第二行输入正整数 M

【数据规模】
1≤N≤10^100
1≤M≤2^31-1

输出描述:

输出到标准输出。
对于每组数据,输出一行:
余数 S 能分解出的不同质因数的个数。
示例1

输入

2
68
40
6
180

输出

2
2

分析:现在题目都好考思维了,求余数简单,题目就是求s有多少素数,它可以整除.

南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解
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 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FFI(a, b) for(int i = a; i < b; i++)
 76 #define FFJ(a, b) for(int j = a; j < b; j++)
 77 #define RR(a, b) for(int i = a; i > b; i++)
 78 #define ME(a, b) memset(a, b, sizeof(a))
 79 #define SC(x) scanf("%d", &x)
 80 #define PR(x) printf("%d\n", x)
 81 #define INF 0x3f3f3f3f
 82 #define MAX 1001
 83 #define MOD 1000000007
 84 #define E 2.71828182845
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 const double PI = acos(-1.0);
 89 typedef pair<int, int> Author;
 90 vector<pair<string, int> > VP; 
 91 
 92 int main(void){
 93     #ifdef LOCAL
 94         freopen("in.txt", "r", stdin);
 95         freopen("out.txt", "w", stdout);
 96     #endif
 97     ios::sync_with_stdio(false); cin.tie(0);
 98     LL T, len, n, m, n1, res, ans;string str;
 99     
100     cin>>T;
101     while(T--){
102         cin>>str>>m;n = res = 0;
103         FFI(0, str.size())n = (n * 10 + str[i] - '0') % m;
104         n1 = (LL)sqrt(n);
105         
106         //ÖÊÒòÊý
107         FFI(2, n1 + 1)if(n % i == 0){res++;while(n % i == 0) n /= i;}
108         
109         if(n != 1)res++;
110         cout<<res<<endl;
111     }
112     
113     return EXIT_SUCCESS;
114 }
View Code

链接:https://www.nowcoder.com/acm/contest/122/G
来源:牛客网

阿汤的数组
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

阿汤同学为了准备下学期的 ACM-ICPC,刷了很多的题目,他觉得自己已经比较厉害了,于是想出个题目考考你。现在他给你一个数组 A,问你是否能将该数组划分成数组 B、C 使得 B 数组的平均数和C 数组的平均数相等,数组 B 和 C 都不能为空。

输入描述:

从标准输入读入数据。
输入包含多组数据,第一行一个整数 T 代表数据组数。接下来依次描述每组数据,对于每组数据:
第一行输入正整数 N,第二行输入 N 个非负整数

1≤|A|≤30 (数组 A 的长度范围在 1 到 30 之间 )
0≤A[i]≤10000 (数组 A 中的元素)

输出描述:

输出到标准输出。
对于每组数据,输出一行:
如果能划分成满足题目要求的数组 B 和 C 则输出 yes,否则输出no
示例1

输入

1
8
1 2 3 4 5 6 7 8

输出

yes

说明

样例说明:将数组 A 划分成【1,4,5,8】和【2,3,6,7】,平均数为 4.5

分析:DP,板子题

南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 int t,n;
 5 bool dp[310005][33];
 6 int s[35];
 7 int solve(){
 8     cin>>n;
 9     double sum=0;
10     for(int i=1;i<=n;i++){
11         cin>>s[i];
12         sum+=s[i];
13     }
14     sum/=n;
15     memset(dp,0,sizeof(dp));
16     int maxn=0;
17     dp[0][0]=1;
18     for(int i=1;i<=n;i++){
19         for(int j=maxn;j>=0;j--){
20             for(int z=0;z<n-1;z++){
21                 if(dp[j][z]){
22                     int k=j+s[i];
23                     dp[k][z+1]=1;
24                     maxn=max(maxn,k);
25                     if(fabs(k*1.0/(z+1)-sum)<=0.0000000001){
26                     //  cout<<k<<" "<<z+1<<endl;
27                     //  cout<<sum<<endl;
28                         return 1;
29                     }
30                 }
31             }
32         }
33     }
34     return 0;
35 }
36 int main()
37 {
38     int t;
39     cin>>t;
40     while(t--){
41         if(solve()) cout<<"yes"<<endl;
42         else cout<<"no"<<endl;
43     }
44     return 0;
45 }
View Code

链接:https://www.nowcoder.com/acm/contest/122/H
来源:牛客网

小q的数列
时间限制:C/C++ 1秒,其他语言2秒
空间限制:C/C++ 32768K,其他语言65536K
64bit IO Format: %lld

题目描述

小q最近迷上了各种好玩的数列,这天,他发现了一个有趣的数列,其递推公式如下:

f[0]=0 f[1]=1;
f[i]=f[i/2]+f[i%2];(i>=2)

现在,他想考考你,问:给你一个n,代表数列的第n项,你能不能马上说出f[n]的值是多少,以及f[n]所代表的值第一次出现在数列的哪一项中?(这里的意思是:可以发现这个数列里某几项的值是可能相等的,则存在这样一个关系f[n'] = f[n] = f[x/2]+f[x%2] = f[x]...(n'<n<x) 他们的值都相等,这里需要你输出最小的那个n'的值)(n<10^18)

输入描述:

输入第一行一个t
随后t行,每行一个数n,代表你需要求数列的第n项,和相应的n'
(t<4*10^5)

输出描述:

输出每行两个正整数
f[n]和n',以空格分隔
示例1

输入

2
0
1

输出

0 0
1 1

分析:这题我本来也是想找循环环,但是发现有点问题,后来再用n / 2 ,n % 2想的

南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解南昌大学航天杯第二届程序设计竞赛校赛网络同步赛题解
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 68                                                                   .:J: :;  .,:K6BQS7:.,.,.
 69                                                                     :r7Ji;r7;::;;.
 70                                                                         .                                                                             **/
 71                                                                         
 72 #include <bits/stdc++.h>
 73 using namespace std;
 74 
 75 #define FFI(a, b) for(int i = a; i < b; i++)
 76 #define FFJ(a, b) for(int j = a; j < b; j++)
 77 #define RR(a, b) for(int i = a; i > b; i++)
 78 #define ME(a, b) memset(a, b, sizeof(a))
 79 #define SC(x) scanf("%d", &x)
 80 #define PR(x) printf("%d\n", x)
 81 #define INF 0x3f3f3f3f
 82 #define MAX 1001
 83 #define MOD 1000000007
 84 #define E 2.71828182845
 85 #define M 8
 86 #define N 6
 87 typedef long long LL;
 88 const double PI = acos(-1.0);
 89 typedef pair<int, int> Author;
 90 vector<pair<string, int> > VP; 
 91 
 92 int main(void){
 93 //    #ifdef LOCAL
 94 //        freopen("in.txt", "r", stdin);
 95 //        freopen("out.txt", "w", stdout);
 96 //    #endif
 97 //    ios::sync_with_stdio(false); cin.tie(0);
 98     LL T, n, ans, len;
 99     
100     cin>>T;
101     while(T--){
102         scanf("%lld", &n); len = ans = 0;
103         while(n > 1){
104             if(n % 2 == 1) len++;
105             n /= 2;
106         }
107         len += n == 0 ? 0 : 1;
108         FFI(0, len) ans = ans * 2 + 1;
109         printf("%lld %lld\n", len, ans);
110     }
111     
112     return EXIT_SUCCESS;
113 }
View Code