Fzu sort l

/*************************************** * ****** <Br/> ** description: 1 ~ An arrangement of N (<= 50000), which counts the number of consecutive subsequences with an odd number in length. The median is B. <Br/> median refers to the number in the center after all elements are arranged in ascending order. <Br/> ** sample input: 1 // test cases <br/> 7 4 // n B <br/> 5 7 2 4 3 1 6 <br/> ** sample output: 4 // {4}, {7, 2, 4}, {5, 7, 2, 3}, and {5, 7, 2, 4, 3, 6} <br/> ** algorithm: mathematical + bipartite enumeration <br/> ** analysis: 1.a[ I] = SGN (C [I]-B ), s [I] = A [1] + A [2] +... + A [I] <br/> 2. the original problem is equivalent to finding s [a] = s [B], where 0 <= A <C <= B <= n, c [c] = B <br/> 3. so you can sort 0--c-1 and C--N, and then two points to find the number of the same left and right; <br/> 4. if this time is the same as the previous number, you do not need to repeat it. <br/> ** Note: This is a good question for Binary training! <Br/> ************************************ * *********/<br/> # include <iostream> <br/> # include <stdio. h> <br/> # include <algorithm> <br/> # define maxn 100010 <br/> using namespace STD; </P> <p> INT cases, n, b, I, m, p; <br/> int s [maxn]; <br/> int binary1 (int K) {<br/> int L = p-1, R = n + 1; <br/> while (L + 1 <r) {<br/> int mid = (L + r)/2; <br/> If (s [Mid] <k) L = mid; else r = mid; <br/>}< br/> return l; <br/>}< br/> int binary2 (int K) {<br/> int L = P, R = n + 1; <br/> while (L + 1 <r) {<br/> int mid = (L + r)/2; <br/> If (s [Mid] <= k) L = mid; else r = mid; <br/>}< br/> return R; <br/>}< br/> int main () {<br/> freopen ("B. in "," r ", stdin); <br/> scanf (" % d ", & cases); <br/> while (cases --) {<br/> scanf ("% d", & N, & B); <br/> S [0] = 0; P = 0; <br/> for (I = 1; I <= N; I ++) {<br/> scanf ("% d", & M ); <br/> If (M> B) M = 1; else <br/> If (M <B) M =-1; else <br/> m = 0, P = I; <br/> S [I] = s [I-1] + m; <br/>}</P> <p> sort (S, S + p); <br/> sort (S + P, S + n + 1); </P> <p> int ans = 0, last =-1 <20; <br/> for (I = 0; I <p; I ++) {<br/> If (s [I]> = s [p] & S [I] <= s [N]) {<br/> if (I >=1 & S [I] = s [I-1]) {ans + = last; continue ;} <br/> int k1 = binary1 (s [I]); // meets the requirements of S [k2] <s [I], s [k1 + 1]> = s [I] <br/> int k2 = binary2 (s [I]); // s [k2]> S [I], s [k2-1] <= s [I] <br/> // cout <k1 <"<k2 <Endl; <br/> last = k2-k1-1; <br/> ans + = last; <br/>}< br/> printf ("% d/N", ANS ); <br/>}< br/> return 0; <br/>}< br/>