task1分治

多数元素

class Solution {
public:
int majorityElement(vector& nums) {
int tar = nums[0], times = 1;
for(int i = 1; i < nums.size(); i++){
if(times == 0){
tar = nums[i];
times = 1;
}
else if(nums[i] != tar)
times–;
else
times++;
}

    return tar;
}

};

最大子序和

class Solution {
public:
struct st{
int ls, rs, ms, is;
};

int maxSubArray(vector& nums) {
    return get(nums, 0, nums.size() - 1).ms;
}

st push(st l, st r){
    int is = l.is + r.is;
    int ls = max(l.ls, l.is + r.ls);
    int rs = max(r.rs, r.is + l.rs);
    int ms = max(max(l.ms, r.ms), l.rs + r.ls);
    return st{ls, rs, ms, is};
}

st get(vector ar, int l, int r){
    if(l == r) return st{ar[l], ar[l], ar[l], ar[l]};
    int m = (l + r) >> 1;
    st lsub = get(ar, l, m);
    st rsub = get(ar, m + 1, r);
    return push(lsub, rsub);
}

};

Pow(x, n)

class Solution {
public:
double eps = 1e-8;
bool eq(double x, double y){
return x == y;
}
double myPow(double x, int n) {
// if(eq(x, 0) || eq(x, 1))
// return x;
// if(n == 0)
// return 1;

    int of = 0;
    
    if(n < 0){
        if(n != INT_MIN){
            n = -n;
        }
        else{
            n = INT_MAX;
            of = 1;
        }
        x = 1 / x;
    }


    double ans = 1;
    if(of)
        ans = ans * x;
    while(n){
        if(n & 1)
            ans = ans * x;
        x = x * x;
        n = n >> 1;
    }

    return ans;
}

};