# A few simple numerical algorithms implemented in STL algorithm

Source: Internet
Author: User
Tags printf

Review of the STL source analysis, the first few simple algorithms to achieve their own. The parameter does not use an iterator, directly using an integer array.
#include <cstdlib> #include <cstdio>/* applies to ordered intervals, judging whether the elements in B are all contained in a. The default is ascending, if descending, change the comparison symbol */BOOL includes (int* A, size_t al, int *b, size_t bl) {size_t i = 0, j = 0; while (I < Al && Amp J < BL) {if (A[i] > B[j]) return false, else if (A[i] < b[j]) i++; else i++, j + +;} return j = = BL; }/* Place the sequence A in the previous segment with the pred condition as true, False in the back segment */int* partition (int* A, size_t begin, size_t end, predicate pred) {size_t i = Begin, J = end; while (true) {while (true) {if (i = = j) return I; else if (pred (A[i])) i++; else break;} while (true) {if (i = = j) ret Urn i; else if (!pred (A[j])) j--; else break; }//swap a[i] = A[i] + a[j]; A[J] = A[i]-a[j]; A[i] = A[i]-a[j]; i++; }}/* Overall exchange a[begin, middle-1] and A[middle, end-1] */inline void rotate (int* A, size_t begin, size_t middle, size_t end) { for (size_t i = middle;;) {//swap int temp = a[i]; A[i] = A[begin]; A[begin] = temp; i++, begin++; if (begin = = middle) {if (i = = end) return; MI Ddle = i; } else if (i = = end) i = middle; } */* in A[B1, e1-1] to find the first occurrence of B[B2, e2-1], failed to return e1 */int search (int* A, size_t B1, size_t E1, int* B, size_t B2, size_t E2 {size_t D1 = e1-b1; size_t d2 = e2-b2; if (D1 < D2) return e1; size_t i = b1, j = B2; while (j! = E2) {if (A[i] = = B[j]) i++, j + +; else if (D1 = = D2) return E1; else {i = ++b1; j = B2; d1--;}} return B1; } int main () {int a[] = {1, 2, 3, 4, 5}; int b[] = {1, 3, 5}, if (includes (A, sizeof (a)/sizeof (int), B, sizeof (b)/siz EOF (int))) printf ("yes/n"); int c[] = {1, 2, 3, 4, 5, 6, 7, 8, 9}; Rotate (c, 0, (sizeof (c)/sizeof (int))/2, sizeof (c)/sizeof (int)); for (size_t i = 0; i < sizeof (c)/sizeof (int); i++) printf ("%d", c[i]); printf ("/n"); int d[] = {1, 2, 3, 4, 5}; int e[] = {7, 5}; printf ("%d/n", search (d, 0, sizeof (d)/sizeof (int), E, 0, sizeof (e)/sizeof (int))); return 0; }

Related Keywords:

The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion; products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the content of the page makes you feel confusing, please write us an email, we will handle the problem within 5 days after receiving your email.

If you find any instances of plagiarism from the community, please send an email to: info-contact@alibabacloud.com and provide relevant evidence. A staff member will contact you within 5 working days.

## A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

• #### Sales Support

1 on 1 presale consultation

• #### After-Sales Support

24/7 Technical Support 6 Free Tickets per Quarter Faster Response

• Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.