NOIP幂次方

#include<stdio.h>
int c[] = { ,,,,,,,,,,,,,,, };//由题意n最大为20000，所以最多会用到2的14次方
//为了防止mid+1出错，故写到15次方

int binarySearch(int x, int mid) {
    if (x >= c[mid] && x < c[mid + ]) {
        return mid;
    }
    if (x >= c[mid + ])
        return ;
    return -;
}

int serch(int x) {//利用二分查找找到x的最大二次方
    int left = ;
    int right = ;
    int mid;
    while (left <= right) {
        mid = (left + right) >> ;
        if (binarySearch(x, mid) == -) {
            right = mid - ;
        }
        else if (binarySearch(x, mid) == ) {
            left = mid + ;
        }
        else {
            return mid;
        }
    }
}
void divite(int x) {//分治求解：因为每个整数的划分方法是一样的
    if (x == )
        return;
    int flag = serch(x);
    int li = x - c[flag];
    if (flag == ) {
        printf("2(0)");
    }
    if (flag == ) {
        printf("");
    }
    if (flag > ) {
        printf("2(");
        divite(flag);
        printf(")");
    }
    if (li > ) {
        printf("+");
        divite(li);
    }
}

int main() {
    int n;
    scanf("%d", &n);
    divite(n);
    return ;
}