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 ;

}