(！！！第一次抽獎中衣服，這個題解是必須要寫的！！！)
A,D簽到題就不講了，C題注意題目中2人距離不變，所以時間t = L / v1即可
鏈接：https://www.nowcoder.com/acm/contest/115/B
來源：牛客網
大吉大利
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
給出一個出生日期，比如：1999-09-09，
問：從出生那一天開始起，到今天2018-04-21為止（包括出生日期和今天），有多少天，年月日都不包含數字4？
問：從出生那一天開始起，到今天2018-04-21為止（包括出生日期和今天），有多少天，年月日都不包含數字4？
輸入描述:
第一行輸入一個整數T（表示樣例個數）
接下來T組樣例
每個樣例一行，包含一個字符串“yyyy-mm-dd”（1990<=yyyy<=2018）
題目保證測試數據的正確性
輸出描述:
輸出題意要求的天數
示例1
輸入
1 1999-09-09
輸出
5020
分析：一開始是想直接每年每年算，結果直接崩了，首先看清題意，year從1990開始，所以year % 10 != 4即可,再循環到2018-4-21號，控制月份和年份以及相關條件即可
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鏈接：https://www.nowcoder.com/acm/contest/115/E
來源：牛客網
消息列表
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
當你的好友給你發來一條消息，你的消息列表上就會置頂顯示該好友的名字以及該好友發給你的消息總數，換句話說，你的消息列表里的好友是按跟你發消息的時間進行排序的，給你發消息的時間離當前時間越近的好友將排到越前面。當然，你可能會手動置頂一些好友，那么其他的好友給你發消息時，他們的名字就只能在你手動置頂好友的后面再置頂了。如果消息被你查看或者忽略，又或者你把好友消息刪除了，消息總數將重置為0。
根據用戶的需求，有以下幾個功能，需要你來實現：
（1）recv：收到一條好友消息，對於手動置頂好友的消息，將在“手動置頂好友列表”里置頂；對於其他好友的消息，將在“手動置頂好友列表”之下的消息列表里置頂，同時都需要顯示該好友的消息總數。
（2）view：查看好友消息，將使該好友消息數變為0。
（3）up：手動置頂好友。
（4）down：取消手動置頂。
（5）delete：刪除好友消息，這個操作將使該好友從消息列表中刪除，同時取消對該好友的手動置頂（如果存在的話）。
假設初始消息列表為空，經過了一系列好友消息的操作之后，最終的消息列表將是怎么樣的呢？
根據用戶的需求，有以下幾個功能，需要你來實現：
（1）recv：收到一條好友消息，對於手動置頂好友的消息，將在“手動置頂好友列表”里置頂；對於其他好友的消息，將在“手動置頂好友列表”之下的消息列表里置頂，同時都需要顯示該好友的消息總數。
（2）view：查看好友消息，將使該好友消息數變為0。
（3）up：手動置頂好友。
（4）down：取消手動置頂。
（5）delete：刪除好友消息，這個操作將使該好友從消息列表中刪除，同時取消對該好友的手動置頂（如果存在的話）。
假設初始消息列表為空，經過了一系列好友消息的操作之后，最終的消息列表將是怎么樣的呢？
輸入描述:
第一行輸入一個整數T（表示樣例個數）
接下來T組樣例。
每組樣例
第一行輸入一個整數M，表示操作數（1≤M≤1000000）；
接下來M行，
每行輸入一個操作，由一個操作類型和一個好友id構成，之間以空格分開，操作類型如上面5個英文單詞表示，
例如：“recv 123456”表示接收到id為123456的好友的一條消息，“delete 123456”表示在消息列表中刪除 id 為123456的好友的消息記錄。
為了簡化問題，一開始消息列表為空並假設好友名字id由六位數字“唯一”標識（000000≤id≤999999），
題目保證輸入數據的一致性。
輸出描述:
每組樣例，
輸出最后的消息列表，自頂向下，每行輸出一個：“好友id 消息數”。
每組樣例后空一行。
示例1
輸入
1 13 recv 000001 recv 000002 up 000002 view 000001 recv 000002 recv 000004 up 000004 up 000001 recv 000004 recv 000003 view 000001 view 000004 down 000002
輸出
000004 0 000001 0 000003 1 000002 2
分析:說實話，第一次遇到這個sort類型題目，感覺挺實用的，和QQ消息差不多，加time,flag保證題目中“越接近現在排序越靠前”和“置頂比不置頂牛皮”即可，再就是最后格式輸出，puts("");是AC，cout<<endl；會WA,另外ME(mes, 0)可以初始化結構體且內部變量可以直接++,另外別忽略id為000000即可
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鏈接：https://www.nowcoder.com/acm/contest/115/F
來源：牛客網
解方程
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
對於方程 2018 * x ^ 4 + 21 * x + 5 * x ^ 3 + 5 * x ^ 2 + 14 = Y，
告訴你Y的值，你能找出方程在0~100之間的解嗎？
告訴你Y的值，你能找出方程在0~100之間的解嗎？
輸入描述:
第一行輸入一個正整數T（表示樣例個數）
接下來T組樣例
每組樣例一行，輸入一個實數Y
輸出描述:
一行輸出一個樣例對應的結果，
輸出方程在0~100之間的解，保留小數點后4位小數；如果不存在，輸出 -1
示例1
輸入
2 1 20180421
輸出
-1 9.9993
分析:這題牛皮了，一開始想的是循環，可是發現0~100內可以為小數，二分還是牛逼。
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鏈接：https://www.nowcoder.com/acm/contest/115/G
來源：牛客網
小Q的口袋校園
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
周末，小Q喜歡在PU口袋校園上參加各種活動刷績點，體驗豐富多彩的大學生活。
但是每個活動有各自的開始時間、結束時間、Happy值以及績點數，活動之間可能存在時間沖突。
小Q是一個認真踏實的女孩，她在同一時間只會參加一個活動。
給出每一天的活動數，求小Q在同一天內所能獲得的最大Happy值 與 績點數。
小Q是一個活潑開朗的女孩，她總是尋求最大的Happy值，如果Happy值相同，才會盡可能使績點數更多。
輸入描述:
第一行輸入一個整數T（表示樣例個數）
接下來T組樣例
每組樣例第一行輸入一個整數N（表示有幾項活動）
接下來N行，每行輸入s,e,h,p 4個整數，分別代表一個活動的開始時間、結束時間、Happy值以及績點數
0<=s<e<=24,1<=h,p<=100，1<=T,N<=20
輸出描述:
輸出T行
一行輸出一個樣例對應的結果
小Q在一天內所能獲得的最大 Happy值 與 績點數
示例1
輸入
1 4 1 3 1 1 2 5 2 1 3 7 2 1 5 8 1 2
輸出
3 3
說明
有兩種方案
1）選擇 第1、3兩個活動，可以獲得 Happy值 3 和 績點數 2；
2）選擇 第2、4兩個活動，可以獲得 Happy值 3 和 績點數 3.
分析：這個和活動安排問題有點像，算是加強版了吧，轉移方程我是沒推出來，DFS倒是騷了一波，預處理一下，排個序就可以了
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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 FF(i, a, b) for(int i = a; i < b; i++) 76 #define RR(i, a, b) for(int i = a; i > b; i++) 77 #define ME(a, b) memset(a, b, sizeof(a)) 78 #define SC(x) scanf("%d", &x) 79 #define PR(x) printf("%d\n", x) 80 #define INF 0x3f3f3f3f 81 #define MAX 30 82 #define MOD 1e9+7 83 #define E 2.71828182845 84 #define PI 3.14159265358979 85 #define M 8 86 #define N 6 87 typedef long long LL; 88 typedef pair<int, int> Author; 89 vector<pair<string, int> > VP; 90 91 struct Node{ 92 int b,e,h,p; 93 }acvity[MAX]; 94 int happy, score, n; 95 96 void DFS(int ha, int so, int num, int ee){ 97 if(happy < ha || (happy == ha && score < so)){happy = ha; score = so;} 98 99 if(num >= n) return ; 100 if(acvity[num].b >= ee) DFS(ha + acvity[num].h, so + acvity[num].p, num + 1, acvity[num].e); 101 DFS(ha, so, num + 1, ee); 102 } 103 bool cmp(const Node& a, const Node& b){return a.b < b.b;} 104 105 int main(void){ 106 #ifdef LOCAL 107 freopen("in.txt", "r", stdin); 108 freopen("out.txt", "w", stdout); 109 #endif 110 ios::sync_with_stdio(false); cin.tie(0); 111 int i, j, T; 112 113 cin>>T; 114 while(T--){ 115 cin>>n;happy = score = 0; 116 for(i = 0; i < n; i++)cin>>acvity[i].b>>acvity[i].e>>acvity[i].h>>acvity[i].p; 117 sort(acvity, acvity + n, cmp); 118 DFS(0, 0, 0, 0); 119 cout<<happy<<" "<<score<<endl; 120 } 121 122 return EXIT_SUCCESS; 123 }
鏈接：https://www.nowcoder.com/acm/contest/115/H
來源：牛客網
小P的數學問題
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
晚上，小P喜歡在寢室里一個個靜靜的學習或者思考，享受自由自在的單身生活。
他總是能從所學的知識散發出奇妙的思維。
今天他想到了一個簡單的階乘問題，
0！= 1
1！= 1
2！= 1 * 2 = 2
3！= 1 * 2 * 3 = 6
4！= 1 * 2 * 3 *4 = 24
5！= 1 * 2 * 3 *4 * 5 = 120
。
。
如果 n=1000000000，那么n的階乘會是多少呢，小P當然知道啦，那么你知道嗎？
輸入描述:
第一行輸入一個整數T（表示樣例個數）
接下來T組樣例
每組樣例一行，輸入一個整數N（0<=N<=1000000000）
輸出描述:
輸出T行
每一行輸出N的階乘 N！（由於這個數比較大，所以只要輸出其對1000000007取膜的結果即可）
示例1
輸入
2 0 1000000000
輸出
1 698611116
分析：都對n!情有獨鍾啊，斯特靈公式求n!位數,或者數組模擬求n!（不%），或者大數n!，fft算法或者python,java大法,再就是這個題目了吧
，求n!要%MOD，板子題，沒什么講的
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define MAX 10000000 4 #define MOD 1000000007 5 #define LL long long 6 7 LL ans[120]={1,682498929,491101308,76479948,723816384,67347853,27368307,625544428,199888908, 8 888050723,927880474,281863274,661224977,623534362,970055531,261384175,195888993,66404266, 9 547665832,109838563,933245637,724691727,368925948,268838846,136026497,112390913,135498044, 10 217544623,419363534,500780548,668123525,128487469,30977140,522049725,309058615,386027524, 11 189239124,148528617,940567523,917084264,429277690,996164327,358655417,568392357,780072518, 12 462639908,275105629,909210595,99199382,703397904,733333339,97830135,608823837,256141983, 13 141827977,696628828,637939935,811575797,848924691,131772368,724464507,272814771,326159309, 14 456152084,903466878,92255682,769795511,373745190,606241871,825871994,957939114,435887178, 15 852304035,663307737,375297772,217598709,624148346,671734977,624500515,748510389,203191898, 16 423951674,629786193,672850561,814362881,823845496,116667533,256473217,627655552,245795606, 17 586445753,172114298,193781724,778983779,83868974,315103615, 18 965785236,492741665,377329025,847549272,698611116}; 19 20 21 22 int main(void){ 23 LL i, T, n, m, cnt; 24 25 cin>>T; 26 while(T--){ 27 cin>>n; 28 m = n / MAX; 29 cnt = ans[m]; 30 31 for(i = m * MAX + 1; i <= n; i++)cnt = (cnt * i) % MOD; 32 cout<<cnt<<endl; 33 } 34 35 return EXIT_SUCCESS; 36 }
鏈接：https://www.nowcoder.com/acm/contest/115/I
來源：牛客網
小P和小Q
時間限制：C/C++ 1秒，其他語言2秒
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
空間限制：C/C++ 32768K，其他語言65536K
64bit IO Format: %lld
題目描述
小P和小Q是好朋友，今天他們一起玩一個有趣的游戲。
他們的初始積分都為1，贏的人可以將自己的分數乘以 (K的平方)，而輸的人也能乘以K。
他們玩的太開心了，以至於忘了自己玩了多久，甚至 K 是多少和游戲進行的回合數 N 都忘了。
現在給出他們倆最終的積分a,b，請問是否存在正整數K、N滿足這樣的積分，判斷他們的游戲結果是否可信。
輸入描述:
第一行輸入一個整數T（表示樣例個數）
接下來T組樣例
每組樣例一行，輸入兩個正整數a,b（0<a,b<=1e9）
輸出描述:
輸出T行
一行輸出一個樣例對應的結果
若結果可信，輸出 Yes
否則，輸出 No
示例1
輸入
6 2 4 75 45 8 8 16 16 247 994 1000000000 1000000
輸出
Yes Yes Yes No No Yes
備注:
每回合的K可能不同
分析：這題我感覺是真的有意思，小p：k^(2a) * k^(n - a)；小q:k^(2n - 2a) * k^(a);所以二者相乘k ^ 3n，說明k,n一定為整數即(int)(cbrt(a * b) + 0.5)若不等於(a * b)^3者說明不存在k,n
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define MAX 10000000 4 #define MOD 1000000007 5 #define LL long long 6 7 int main(void){ 8 LL i, a, b, T, ans; 9 10 //由題意推得k,n應為正數 11 cin>>T; 12 while(T--){ 13 cin>>a>>b; 14 ans = (LL)(cbrt(a * b) + 0.5); 15 if(ans * ans * ans == a * b) cout<<"Yes"<<endl; 16 else cout<<"No"<<endl; 17 } 18 19 return EXIT_SUCCESS; 20 }