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
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。
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。
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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
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？
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
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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
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。
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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LBBgGgZHsr;vrr;rcXpP5UUS5X5S2XS5SS1URBggpEZgDDGOGggRD 57 ,;irrs2KgQRPJJJSUXKXSHJpBM . . . ZBBOUsr;:. ,,,rHgZPUaSaUSUSSXaUPXsSBB6ZZp6EEGOOGDEggO 58 ;BBQQZaSJ5J15SJDQ7 .iEBL:. . .,:,:;SRQGZXPXXSHaS2UUSUSswQB6OZgG6pZZGODEOGDE 59 JgUri1aGEpEpXSS5LBQ, .rLap2Jr ;712OO6ZggRDZXHXpKK2211LLLc7wgBa2PODRGE6OZgDgGDGD 60 7QB; ,Jc76DaZXOZgDPEBBX, :c1HEZ1c;:,.,:;cHDgQQggMZgGOZO6OEGpXsLLsJXHPOBBBc;vss5EMggZDODOgED6 61 LZJ::iwrrr72EPgXU5OBBBBBBBp7: .i5P6wL;, .:LS6DMgMgZKgEOPOGGDDGEXULvLSOBBBBBQBgDQDJ5rrr7JOQQDDZGEgDD 62 wp rR5, .HQRX7r72RQBBBBBQBMEK6Uc7sc7cHOU:, .:LHgOQg6ZOPZpZGGEDERZPwcr7LKDDaJr:. rBX;;:r:ir1DBQMgRgDp 63 7D,sw: , Z. .LgBBQBBOsJaQBBBBBBXrr. .:rwZBBQgROgDgDOERRMgROOScrLsPP5r, 7Qs,::::::rUgRgZOG6 64 ,Q57:r ,: ,r U: .;; .vL7L;r:UG7.:sr;JSgQBgDEZXXUSXpHa21svr7r7s2v:. iQK...:,:::;LLJJws 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 ", 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 }
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
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 }
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
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)
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
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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 ", 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 ", len, ans); 110 } 111 112 return EXIT_SUCCESS; 113 }