hdu 4968 Improving the GPA(dp)

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题目链接：hdu 4968 Improving the GPA

题目大意：给定平均分和科目数量，要求保证及格的前提下，求平均绩点的最大值和最小值。

解题思路：官方题解是暴力枚举，我的做法是dp。dp[i][j]表示i个科目，总分j（扣掉基础60分，减少复杂度）的情况,然后预处理出dp数组之后答案是通用的。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 10;const int maxm = 400;const double INF = 0x3f3f3f3f;double l[maxn+5][maxm+5], r[maxn+5][maxm+5];inline double cal (int k) {    if (k < 10)        return 2.0;    else if (k < 15)        return 2.5;    else if (k < 20)        return 3.0;    else if (k < 25)        return 3.5;    else        return 4.0;}void init () {    for (int i = 0; i <= maxn; i++) {        for (int j = 0; j <= maxm; j++) {            l[i][j] = 50;            r[i][j] = 0;        }    }    l[0][0] = r[0][0] = 0;    for (int i = 0; i < maxn; i++) {        for (int j = 0; j <= maxm; j++) {            for (int k = 0; k <= 40; k++) {                if (j + k > maxm)                    break;                l[i+1][j+k] = min(l[i+1][j+k], l[i][j] + cal(k));                r[i+1][j+k] = max(r[i+1][j+k], r[i][j] + cal(k));            }        }    }}int main () {    init();    int cas;    scanf("%d", &cas);    while (cas--) {        int score, n;        scanf("%d%d", &score, &n);        score = (score - 60) * n;        printf("%.4lf %.4lf\n", l[n][score] / n, r[n][score] / n);    }    return 0;}

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