PAT A1103

PAT A1103

标签（空格分隔）： PAT

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解题思路： DFS

#include <cstdio>
#include <vector>
using namespace std;

int n, k, p;
int maxFacSum = -1;         //记录最大因子和
vector<int> ans, temp, fac;

int power(int x) {
    int ans = 1;
    for(int i = 0; i < p; i++) {
        ans *= x;
    }
    return ans;
}

void init() {
    int i = 0, temp = 0;    //初始化fac，把因子的平方存入其中
    while(temp <= n) {
        fac.push_back(temp);
        temp = power(++i);
    }
}

void DFS(int index, int nowK, int sum, int facSum) {
    if(nowK == k && sum == n) {        //dfs终止条件
        if(facSum > maxFacSum) {      //目前的因子和大于最大因子和
            maxFacSum = facSum;      
            ans = temp;               //更新答案vector
        }
        return;
    }
    if(nowK > k || sum > n) return;     //此时不可能满足条件，直接返回
    if(index - 1 >= 0) {              
        temp.push_back(index);
        DFS(index, nowK + 1, sum + fac[index], facSum + index);    //选择index
        temp.pop_back();
        DFS(index - 1, nowK, sum, facSum);    //不选index
    }
}

int main() {
    scanf("%d%d%d", &n, &k, &p);
    init();
    DFS(fac.size() - 1, 0, 0, 0);
    if(maxFacSum == -1) printf("Impossible\n");
    else {
        printf("%d = %d^%d", n, ans[0], p);
        for(int i = 1; i < ans.size(); i++) {
            printf((" + %d^%d"), ans[i], p);
        }
    }
    return 0;
}