[ACM] sdut 2878 Circle (Gaussian deyuan), sdut2878

[ACM] sdut 2878 Circle (Gaussian deyuan), sdut2878

Circle Time Limit: 2000 ms Memory limit: 65536 K any questions? Click Here ^_^ Description: You have been given a circle from 0 to n rows-equal 1. if you are currently at x, you will move to (x release-interval 1) mod n or (x release + interval 1) mod n with equal probability. now we want to know the expected number of steps you need to reach x from 0. enter The first line contains one integer T-the number of test cases. each of the next T lines contains two integers n, divide x (0 rows ≤ limit x limit <limit n Limit ≤ limit 1000) as we mention above. output For each test case. print a single float number-the expected number of steps you need to reach x from 0. the figure is accurate to 4 decimal places. sample Input

33 25 410 5

Sample output

2.00004.000025.0000

Tip: 2014 Shandong fifth ACM College Student Program Design Competition

Solution:

N nodes numbered 0 to n-1 form a ring. Given a number x, calculate the expected number of steps from node 0 to node x. The probability of a node going to both sides is the same, and each step goes to one node.

Gaussian elimination element, n equations, n unknown numbers. set E [p] to the expected number of steps required to go from node p to node x. Then E [x] = 0;

For each node, E [p] = 0.5 * E [P-1] + 0.5 * E [p + 1] + 1, that is,-0.5 * E [P-1] + E [p]-0.5 * E [p + 1] = 1.

Code:

# Include <iostream> # include <cstdio> # include <cstring> # include <string. h >#include <cmath> # include <iomanip> # include <algorithm> using namespace std; // floating-point Gaussian elimination template const double eps = 1e-12; const int maxm = 1000; // m equations, n variables const int maxn = 1000; int m, n; double a [maxm] [maxn + 1]; /// Augmented Matrix bool free_x [maxn]; // determines whether it is an uncertain variable; double x [maxn]; // returns an int sign (double x) {return (x> eps)-(x <-eps);}/** return value:-1 no solution 0 has only one solution> = 1 has multiple solutions, according Free_x determines which solutions are uncertain */int Gauss () {int I, j; int row, col, max_r; m = n; // n equations, for (row = 0, col = 0; row <m & col <n; row ++, col ++) {max_r = row; for (I = row + 1; I <m; I ++) // find the maximum value of all rows in the current column (reduce the error when division is performed) {if (sign (fabs (a [I] [col])-fabs (a [max_r] [col])> 0) max_r = I;} if (max_r! = Row) {for (j = row; j <n + 1; j ++) swap (a [max_r] [j], a [row] [j]);} if (sign (a [row] [col]) = 0) // all the rows below the current column row are 0 (including row rows) {row --; continue ;} for (I = row + 1; I <m; I ++) {if (sign (a [I] [col]) = 0) continue; double tmp = a [I] [col]/a [row] [col]; for (j = col; j <n + 1; j ++) a [I] [j]-= a [row] [j] * tmp ;}} for (I = row; I <m; I ++) /// col = n 0... 0, a, no solution {if (sign (a [I] [col]) return-1;} if (row <n) // 0... 0, 0, there are multiple solutions, the number of free variable is n-row {for (I = row- 1; I> = 0; I --) {int free_num = 0; // The number of free variable elements int free_index; // The number of free variable elements for (j = 0; j <n; j ++) {if (sign (a [I] [j])! = 0 & free_x [j]) free_num ++, free_index = j;} if (free_num> 1) continue; // The number of uncertain variable values in this row exceeds 1, unable to solve, they are still uncertain variable elements // there is only one uncertain variable element free_index, which can be solved, and the variable is determined by double tmp = a [I] [n]; for (j = 0; j <n; j ++) {if (sign (a [I] [j])! = 0 & j! = Free_index) tmp-= a [I] [j] * x [j];} x [free_index] = tmp/a [I] [free_index]; free_x [free_index] = false;} return n-row;} // there is only one solution, and the strict upper triangle matrix (n = m) for (I = n-1; i> = 0; I --) {double tmp = a [I] [n]; for (j = I + 1; j <n; j ++) if (sign (a [I] [j])! = 0) tmp-= a [I] [j] * x [j]; x [I] = tmp/a [I] [I];} return 0 ;} /// template end int t, xx; int main () {cin> t; while (t --) {cin> n> xx; memset (a, 0, sizeof (a); for (int I = 0; I <n; I ++) {if (I = xx) {a [I] [I] = 1; a [I] [n] = 0; continue;} a [I] [I] = 1; a [I] [n] = 1; a [I] [(I-1 + n) % n] =-0.5; a [I] [(I + 1) % n] =-0.5;} Gauss (); cout <setiosflags (ios: fixed) <setprecision (4) <x [0] <endl;} return 0 ;}