心累，昨天一天3个比赛，全部炸掉，这个陕西师范的比赛我真的读不懂题目，只搞出来7题，难受
A,B,F签到题，不讲了，只讲C,D,I,K
链接：https://www.nowcoder.com/acm/contest/121/C
来源：牛客网
iko和她的糖
时间限制：C/C++ 1秒，其他语言2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
iko超级超级喜欢吃糖，有一天iko想出去玩，她计划从1点走到N点（按1,2,3，...，n的顺序走），每个点都有一个补给站，第i点的补给站有a[i]颗糖，从i点走到i+1点会消耗掉b[i]颗糖，iko在出游的途中可以选择三个补给站，iko想知道她走完全程到达N点时口袋里最多还能剩下几颗糖（初始时iko的口袋里一颗糖都没有）。
输入描述:
第一行输入N（3<=N<=1000）
第二行输入N个数代表a[1].......a[N] (0<=a[i]<=1000 )
第三行输入N-1个数代表b[1]......b[N-1] ( 1<=b[i]<=1000 )
输出描述:
输出一个数字表示iko到达n点时口袋里最多剩下的糖，
若不能到达N点输出-1。
示例1
输入
3 1 3 4 3 4
输出
-1
示例2
输入
5 3 4 5 2 4 3 2 2 2
输出
3
分析：DP题目，建立一个二维数组DP[MAX][4],代表i点,领取j个补给站后的糖果数,和二维0-1背包差不多，2重循环，每一个i，对应3条路，每到一个i站，就要把之前糖果数全部计算出来。
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链接：https://www.nowcoder.com/acm/contest/121/D
来源：牛客网
ZQ的睡前故事
时间限制：C/C++ 1秒，其他语言2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
ZQ是一个拥有n女朋友的万人迷，她的每一个女朋友每天晚上都会挨个给他打电话，要他讲了睡前故事才能睡觉。可是，每次他的女朋友都会挑他在吃鸡的时候打电话，ZQ总是因为挂机被舍友赶出宿舍，于是，ZQ告诉他的女朋友们，别打电话了，他会主动打过去给他们讲故事，再打电话就分手！
于是，ZQ把他的女朋友名字写在纸上，画成一圈，顺时针编号为1~n，然后从1开始顺时针数。在每一次数数中，ZQ数k个就停下来，然后给选中的女朋友打电话讲故事。
现在需要你按顺序告诉我们他给女朋友打电话的顺序输入描述:
先输入一个t，然后t组数据，每行包含两个数字n,k，n<20,k>0
输出描述:
按顺序输出每轮被选中的女朋友的编号。
示例1
输入
3 10 3 5 2 11 4
输出
3 6 9 2 7 1 8 5 10 4 2 4 1 5 3 4 8 1 6 11 7 3 2 5 10 9
分析：orz，这题和CFdiv3#481的C题比较类似，一个循环无限遍历1 ~ n + 1，a==k就归零，保存i，标记一下当前位置，考思维。
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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 FF(a, b) for(int i = a; i < b; i++) 76 #define RR(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 1001 82 #define MOD 1000000007 83 #define E 2.71828182845 84 #define M 8 85 #define N 6 86 typedef long long LL; 87 const double PI = acos(-1.0); 88 typedef pair<int, int> Author; 89 vector<pair<string, int> > VP; 90 91 int main(void){ 92 #ifdef LOCAL 93 freopen("in.txt", "r", stdin); 94 freopen("out.txt", "w", stdout); 95 #endif 96 ios::sync_with_stdio(false); cin.tie(0); 97 int k, T, n, a, b, mark[MAX], ans[MAX]; 98 99 cin>>T; 100 while(T--){ 101 cin>>n>>k;b = 1;ME(mark, 0); 102 FF(1, n + 2){ //无限循环 103 if(i == n + 1) i = 1; 104 if(!mark[i]) a++; 105 if(a == k){ 106 a = 0; 107 ans[b++] = i; 108 mark[i] = 1; 109 } 110 if(b == n + 1) break; 111 } 112 FF(1, b)i == b - 1 ? printf("%d\n", ans[i]) : printf("%d ", ans[i]); 113 } 114 115 return EXIT_SUCCESS; 116 }
链接：https://www.nowcoder.com/acm/contest/121/I
来源：牛客网
排队排队排队
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
ACM竞赛队内要开运动会啦！！！！
竞赛队内的一群阳光乐观积极的队员们迅速的在操场上站成了一支队伍准备开始热身运动。但教练看了一眼觉得队伍高高低低很不整齐。
教练想让大家从低到高站好，每次教练可以任选择一个人令他走到队首，教练想知道他要最少要进行几次这样的操作才能把队伍按从低到高排整齐(身高最低的人站在队首)。
输入描述:
第一行输入N（N<=1000）代表一共有N个队员
第二行输入N个数表示初始时的队伍所对应的每个人的身高（100<=身高<=300）（第一个输入的是队首，最后一个输入的是队尾）
输出描述:
输出教练所需要的最小操作步数
示例1
输入
2 183 185
输出
0
示例2
输入
3 173 186 166
输出
1
说明
一次操作将站在队尾身高为166的队员放在队首就形成了166 173 186的从低到高有序队伍