leetcode3250. 单调数组对的数目 I，仅需1s

题目：
https://leetcode.cn/problems/find-the-count-of-monotonic-pairs-i/description/
不为别的，只是记录下这个超过100%，而且比原先最快的快了一个量级

不知道咋分析，反正得出结论就是，变大不变，变小跟着变，最后一个数，如果是负数，直接为0，如果为正数，就是C(len+n取n)
两个解题时间如图：
1、使用加法计算C

class Solution {public int countOfPairs(int[] nums) {int jj = 0;int len = nums.length;for (int i = 1; i < len; i++) {if (nums[i] - jj >= nums[i - 1]) {jj = nums[i] - nums[i - 1];nums[i] = nums[i - 1];} else {nums[i] -= jj;if (nums[i] < 0) {return 0;}}}return getAns(nums.length, nums[len - 1]);}//使用加法计算private int getAns(int length, int num) {long[] anss = new long[num + 1];for (int i = 0; i < anss.length; i++) {anss[i] = 1;}for (int i = 0; i < length; i++) {for (int j = 1; j <= num; j++) {anss[j] += anss[j - 1];anss[j]%=1_000_000_007;}}return (int) anss[num];}
}

2、使用计算式计算C，因为存在1000_000_007,所以先约分

class Solution {public int countOfPairs(int[] nums) {int jj = 0;int len = nums.length;for (int i = 1; i < len; i++) {if (nums[i] - jj >= nums[i - 1]) {jj = nums[i] - nums[i - 1];nums[i] = nums[i - 1];} else {nums[i] -= jj;if (nums[i] < 0) {return 0;}}}return getAns(nums.length, nums[len - 1]);}private int getAns(int length, int num) {if (length <= num) {return getAA(length, length + num);} else {return getAA(num, length + num);}}//以下计算C(length取n)private int getAA(int n, int length) {if (n == 0) {return 1;}int[] fenzi = new int[n];for (int i = 0; i < n; i++) {fenzi[i] = length - i;}for (int i = 1; i <= n; i++) {int temp = i;int j = 0;while (temp > 1) {int yue = getYue(temp, fenzi[j]);temp /= yue;fenzi[j] /= yue;j++;}}long ans = 1;for (int i : fenzi) {ans *= i;ans %= 1000_000_007;}return (int) ans;}private int getYue(int t1, int t2) {if (t1 == t2) {return t1;}if (t1 > t2) {return getYYY(t1, t2);} else {return getYYY(t2, t1);}}private int getYYY(int t1, int t2) {int t3 = t1 % t2;if (t3 == 0) {return t2;}return getYYY(t2, t3);}
}