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

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。

  1 /**
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 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 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

  1 /**
  2  
  3                                                                        :;LaEaHKEEGpPXU7;,
  4                                                                   .:75pKH11252U252XapZgRQgD6XJscLr;,.
  5                                                                :LXpRgGaX521JLw1JswJJsJs22XHPPEZEGDOMMRDOa7.
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 10                                                 :UBQOKPK6ZOOOEDEO6GZE6EpEpDgDBR:   UBpXHa5aSaUS5SUS5XapPHJc7rrv7rr7sgBBs                           .g.
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 15                                      ,SRJ7sSHGggEOZOEG6OODpOZgggOQQBBBBBQBBQ.,;;. LBOgOgDRDDZODMgQRgRDaa552a252UUa25w5UaU2sLvccs7r7sJZBBMr       ,XQJ:
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 20                                JZ:;rrrc5EPHp6XgpRBBBE;     :i:;;;;;:;;;:c7::r7;,::::::::rv:,:::::::::::,:,:,:,,,,            .;sORQRGX21wsXU:  .... .
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 26                              OX:irsXgQZ:.   .,,:::,,:::,::::::::;;;::r5;r7:           :;;;:7L ..,.,;:,:,:,,,:,:,,,,......        .rU6w;..............
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 58                                        ;BBQQZaSJ5J15SJDQ7                                    .iEBL:.  . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE
 59                                       JgUri1aGEpEpXSS5LBQ,                              .rLap2Jr     ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD
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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.

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 65                                     OBHs;.  :;;,v  H:             :6X;       rpL;7pSrrr7r;::,....,,:;iirrvvvr,                       :BBPs;;:,,,,::;::
 66                                     ::rPDEL:..7:c;7r              pK  .:      :KL.:r:..     ..:icsS5sr:,.                              JgGgBQDOSJss77r
 67                                          :r;:::  ,                PH r,       ,;5: ,::::;7Ls7Lc7;:                                         ,:7JP17rJUs
 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

 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

  1 /**
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 67                                          :r;:::  ,                PH r,       ,;5: ,::::;7Ls7Lc7;:                                         ,:7JP17rJUs
 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

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转载自www.cnblogs.com/meditation5201314/p/9064254.html