对prob 1007的算法改进！

算逆序数可以在O(n)时间内完成：

int count_inver(char *str, int len)
{
        int i;
        int cnt = 0;
        int a[4] = {0};
        for(i = len - 1; i >= 0; i--) {
                switch (str[i]) {
                        case 'A':
                                a[1]++;
                                a[2]++;
                                a[3]++;
                                break;
                        case 'C':
                                a[2]++;
                                a[3]++;
                                cnt += a[1];
                                break;
                        case 'G':
                                a[3]++;
                                cnt += a[2];
                                break;
                        case 'T':
                                cnt += a[3];
                }
        }
        return cnt;
}