Although there is no solution, but the general idea is still there, to ask for AC code.
Problem Descriptionobserve this sequence:
1 3 0 2-1 1-2 ...
The last item in this series is always 2 or 3 less than the previous one.
The building is curious about the series, and he wants to know how many of the integers of length n and S and the latter always add a to the previous item or decrease B. Input Formatthe first line of input contains four integers, n s a B, meaning as stated earlier. output FormatThe output line contains an integer that represents the number of scenarios that satisfy the condition. Since this number is large, the remainder of the output scheme number is divided by 100000007. Sample Input4 2 3Sample Output2Sample Descriptionthe two series are 2 4 1 3 and 7 4 1-2 respectively. data size and conventionsfor 10% of data,1<=n<=5,0<=s<=5,1<=a,b<=5;
For 30% of data,1<=n<=30,0<=s<=30,1<=a,b<=30;
For 50% of data,1<=n<=50,0<=s<=50,1<=a,b<=50;
For 70% of data, 1<=n<=100,0<=s<=500,1<=a, b<=50;
For 100% of data, 1<=n<=1000,-1,000,000,000<=s<=1,000,000,000,1<=a, b<=1,000,000. idea: Recursive enumeration of each number of series, judging and whether or not equal to the input. is the word Count +1. first solve the problem of the range of the first number. The approximate interval is s-n*a;s+n*b;because the run timed out. Further optimization was done. set the first number to X, thetwo enumeration intervals for a sequence:x,x+a,x+2a,x+3a........x+ (n-1) A;x,x-b,x-2b,x-3b.........x-(n-1) b;because and for S,Solution to:x = (s-a*n* (n-1)/2)/n;x = (s + b*n* (n-1)/2)/n;recursive enumeration for the For loop. Timeout Code:
Import Java.util.scanner;class main{static int ans = 0,sum = 0,n,s,a,b;public static void Main (string[] arge) {Scanner cin = new Scanner (system.in); n = Cin.nextint (); s = Cin.nextint (); a = Cin.nextint (); b = Cin.nextint (); for (int i = (s-a*n * (n-1)/2)/n;i<= (S + b*n* (n-1)/2)/n;i++) {DFS (i,n-1);} Ans%= 100000007;sop1 (ans); public static void Dfs (int i,int n) {sum + = i;if (n = = 0) {if (sum = = s) ans++;sum-= I;return;} DFS (i + a,n-1);d FS (i-b,n-1), sum-= i;} public static void Sop1 (Object obj) {System.out.println (obj);}}
commit, or timeout. Really do not know how to optimize the. The AC code is obtained.
Blue Bridge series of waves (unresolved: Run timeout)
