2022-09-05：作为国王的统治者，你有一支巫师军队听你指挥。 : 给你一个下标从 0 开始的整数数组 strength ， 其中 strength[i] 表示第 i 位巫师的力量值。 对于连续的一

fn main() {    let mut arr = vec![1, 3, 1, 2];    let ans = total_strength(&mut arr);    println!("ans = {}", ans);}
const mod0: i64 = 1000000007;
fn total_strength(arr: &mut Vec<i32>) -> i32 {    let n = arr.len() as i32;    let mut pre_sum = arr[0] as i64;    let mut sum_sum: Vec<i64> = vec![];    for _ in 0..n {        sum_sum.push(0);    }    sum_sum[0] = arr[0] as i64;    for i in 1..n {        pre_sum += arr[i as usize] as i64;        sum_sum[i as usize] = (sum_sum[(i - 1) as usize] + pre_sum) % mod0;    }    let mut stack: Vec<i32> = vec![];    for _ in 0..n {        stack.push(0);    }    let mut size: i32 = 0;    let mut ans: i64 = 0;    for i in 0..n {        while size > 0 && arr[stack[(size - 1) as usize] as usize] >= arr[i as usize] {            size -= 1;            let m = stack[size as usize];            let l = if size > 0 {                stack[(size - 1) as usize]            } else {                -1            };            // l（<当前值，且最近，到不了）        m(当前数，做为最小值)      i(<=当前数，到不了的！)            ans += magic_sum(arr, &mut sum_sum, l, m, i);            ans %= mod0;        }        stack[size as usize] = i;        size += 1;    }    while size > 0 {        size -= 1;        let m = stack[size as usize];        let l = if size > 0 {            stack[(size - 1) as usize]        } else {            -1        };        ans += magic_sum(arr, &mut sum_sum, l, m, n);        ans %= mod0;    }    ans as i32}
fn magic_sum(arr: &mut Vec<i32>, sum_sum: &mut Vec<i64>, l: i32, m: i32, r: i32) -> i64 {    let left = (m as i64 - l as i64)        * (sum_sum[(r - 1) as usize]            - if m - 1 >= 0 {                sum_sum[(m - 1) as usize]            } else {                0            }            + mod0)        % mod0;    let right = (r as i64 - m as i64)        * (if m - 1 >= 0 {            sum_sum[(m - 1) as usize]        } else {            0        } - if l - 1 >= 0 {            sum_sum[(l - 1) as usize]        } else {            0        } + mod0)        % mod0;    return arr[m as usize] as i64 * ((left - right + mod0) % mod0);}