1103 Integer Factorization (30分) PAT

在进行序列选择的时候,加入对字典序的比较反而会导致其中一个测试样例出错,注释掉相关内容后全部通过,如下:

#include<cstdio>
#include<cmath>
#include<vector>

const int maxn = 410;

using namespace std;

vector<int> temp, out;
int power[maxn];
int maxsum = 0;

bool test(vector<int> temp, vector<int> out, int K) {
    int sum1 = 0;
    for (int i = 0; i < K; i++) {
        sum1 += temp[i];
    }
    if (sum1 > maxsum) {
        maxsum = sum1;
        return true;
    }
    /*
    else if (sum1 < maxsum) {
        return false;
    }
    else {
        for (int i = 0; i < K; i++) {
            if (temp[i] > out[i]) return true;
        }
    }
    */
    return false;
}

void dfs(int index, int sum, int N, int count, int K) {
    if (count == K && sum == N) {
/*
            for (int i = 0; i < K; i++) {
                printf(" %d^", temp[i]);
                if (i != K - 1) {
                    printf(" +");
                }
            }
            printf("\n");
*/

        if(test(temp, out, K)) {
            out = temp;
        }
        return;
    }

    if (count == K || index == 0 || sum >= N) return;
    temp.push_back(index);
    dfs(index, sum + power[index - 1], N, count + 1, K);
    temp.pop_back();
    dfs(index - 1, sum, N, count, K);
    return;
}

int main() {
    int N, K, P;
    scanf("%d %d %d", &N, &K, &P);
    int i = 1;
    double temp = pow((double)i, P);
    for (i = 1; temp <= N && i < maxn; i++) {
        power[i-1] = (int)temp;
        temp = pow((double)i + 1, P);
    }
    int length = i - 1;
    dfs(length, 0, N, 0, K);
    if (out.size() == 0) {
        printf("Impossible\n");
        return 0;
    }
    printf("%d =", N);
    for (i = 0; i < K; i++) {
        printf(" %d^%d", out[i], P);
        if (i != K - 1) {
            printf(" +");
        }
    }
    printf("\n");

    return 0;
}
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转载自blog.csdn.net/Ike_Lin/article/details/104478733