Problem You are waiting in a long line to get a haircut at a trendy barber shop.
The shop has B barbers on duty, and they are numbered 1 through B.
It always takes the kth barber exactly Mk minutes to cut a customer's hair,
and a barber can only cut one customer's hair at a time. Once a barber
finishes cutting hair, he is immediately free to help another customer. While the shop is open, the customer at the head of the queue always goes
to the lowest-numbered barber who is available. When no barber is available,
that customer waits until at least one becomes available. You are the Nth person in line, and the shop has just opened. Which barber
will cut your hair? Input The first line of the input gives the number of test cases, T. T test cases
follow; each consists of two lines. The first contains two space-separated
integers B and N -- the number of barbers and your place in line.
The customer at the head of the line is number 1, the next one is number 2,
and so on. The second line contains M1, M2, ..., MB. Output For each test case, output one line containing "Case #x: y",
where x is the test case number (starting from 1) and y is the number of the barber who will cut your hair. Limits 1 ≤ T ≤ 100. 1 ≤ N ≤ 109. Small dataset 1 ≤ B ≤ 5. 1 ≤ Mk ≤ 25. Large dataset 1 ≤ B ≤ 1000. 1 ≤ Mk ≤ 100000. Sample Input Output 3 2 4 10 5 3 12 7 7 7 3 8 4 2 1 Case #1: 1 Case #2: 3 Case #3: 1 In Case #1, you are the fourth person in line, and barbers 1 and 2 take 10 and 5 minutes,
respectively, to cut hair. When the shop opens, the first customer immediately has the
choice of barbers 1 and 2, and she will choose the lowest-numbered barber, 1. The second
customer will immediately be served by barber 2. The third customer will wait since there
are no more free barbers. After 5 minutes, barber 2 will finish cutting the second
customer's hair, and will serve the third customer. After 10 minutes, both barbers 1 and
2 will finish; you are next in line, and you will have the choice of barbers 1 and 2, and will choose 1.
题目很简短，也很好理解。
如果用最朴素的方法，不管是暴力还是用 STL 里的 SET ，都需要从 1 遍历到 N （10^9） 那么肯定是不行的
我想到了既然人数那么多说不定有规律可循，过了小数据， TLE 了大数据... 【姿势不对啊
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <fstream> #include <cstring> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #include <list> #include <queue> #include <vector> #include <algorithm> #define Max(a,b) (((a) > (b)) ? (a) : (b)) #define Min(a,b) (((a) < (b)) ? (a) : (b)) #define Abs(x) (((x) > 0) ? (x) : (-(x))) #define MOD 1000000007 #define pi acos(-1.0) using namespace std; typedef long long ll ; typedef unsigned long long ull ; typedef unsigned int uint ; typedef unsigned char uchar ; template<class T> inline void checkmin(T &a,T b){if(a>b) a=b;} template<class T> inline void checkmax(T &a,T b){if(a<b) a=b;} const double eps = 1e-7 ; const int M = 1100011*2 ; const ll P = 10000000097ll ; const int MAXN = 10900000 ; const int INF = 0x3f3f3f3f ; int B, N; int a[1200], cur[1200]; int ans[1000000]; int main(){ std::ios::sync_with_stdio(false); int i, j, t, k, u, v, numCase = 0; ofstream fout ("B-large-practice.out"); ifstream fin ("B-large-practice.in"); fin >> t; int T = t; for (T = 1; T <= t; ++T) { cout << T << endl; fin >> B >> N; for (i = 1; i <= B; ++i) { fin >> a[i]; } memset (cur, 0, sizeof (cur)); int flag = 1; int sum = 0; for (i = 1; i < N; ++i) { cur[flag] += a[flag]; ans[sum++] = flag; int MIN = INF; for (j = 1; j <= B; ++j) { if (cur[j] < MIN) { MIN = cur[j]; flag = j; } } for (j = 1; j < B; ++j) { if (cur[j] ^ cur[j + 1]) { break; } } if (j == B) { break; } } if (i == N) { flag = 1; for (i = 2; i <= B; ++i) { if (cur[i] < cur[flag]) { flag = i; } } } else { if (N % sum == 0) { flag = ans[sum - 1]; } else { flag = ans[N % sum - 1]; } } fout << "Case #" << ++numCase << ": " << flag << endl; } return 0; }