寻找最长递增子序列

思路：

动态规划解题，记当前最长递增子序列的最后一个数为a[i]，长度为c[i];

记a[j]为a[0]到a[i]中所有以j结尾的递增子序列，如果a[i+1].>a[i],，则c[i+1] = MAX(c[i+1],c[j]+1);

#include<stdlib.h>int a[8] = {1,2,-2,3,2,3,4,5}; int c[100] = {1};int max(int a, int b){return a > b ? a : b;}int findMaxSub(int* a, int n){int i, j;for(i = 0; i < n; i++){for(j = 0; j < i; j++){if(a[i] > a[j]){c[i] = max(c[i],c[j] + 1);}}}printf("%d",c[n-1]);return c[n-1];}int main(){findMaxSub(a,8);}