题解征集｜发9.7商汤科技题解，攒人品，赢牛客T恤！

def readData(): y_true, y_pre = [], [] N = int(input()) for _ in range(N): y, y1 = input().split(' ') y = int(y) y1 = float(y1) y_true.append(y) y_pre.append(y1) return N, y_true, y_pre def calTP(t, p, threshold): # y_true = np.array(y_true) # y_pre = np.array(y_pre) TP, FN, FP = 0, 0, 0 for i in range(len(t)): if p[i] >= threshold and t[i] == 1: TP += 1 if p[i] >= threshold: p[i] = 1 else: p[i] = 0 # TP = np.sum(y_true[y_pre >= threshold]) # FN = np.sum(y_true) - TP FN = sum(t) - TP FP = sum(p) - TP return TP, FN, FP def calPR(y_true, y_pre): tmp = y_pre tmp = sorted(tmp) r = [0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9] res = [0, 0, 0, 0, 0, 0, 0] for i in range(len(tmp)): y_t = y_true[:] y_p = y_pre[:] TP, FN, FP = calTP(y_t, y_p, tmp[i]) if (TP+FN) > 0 and (TP+FP) > 0: R = (TP) / (TP + FN) if R in r: P = (TP) / (TP + FP) res[r.index(R)] = int(P*100) for num in res: print(num) if __name__ == '__main__': N, y_true, y_pre = readData() calPR(y_true, y_pre) 本地测试可以，提交之后提示越界，请大佬帮忙指出错误。谢谢

//怪不得选择填空这么少也不太难，原来编程这么难做的么【泪流满面】 //正确率死活只有15%只知道倒着思考，按上楼梯的最后一阶往下看，哪位大佬解解惑【哭哭】 #include <iostream> #include<vector>   usingnamespacestd;   intmain() {          int n, m;         cin >> n >> m;          int*a = new int[n+1];          a[1] = 1;          a[2] = 2;          a[3] = 4;          if(n>3)          {                  for(int i = 4; i <= n; i++)                 {                       for(int j = 0; j<m; j++)                       {                                a[i] = a[i] + a[i - 1 - j];                        }                 }         }        cout << a[n];         return 0; }

我讨厌递归，，，垃圾递归，，，， m = int(strlist[1]) print(dancestep(n,m) ps:为什么代码会乱码？

跳出了5%台阶还有机会吗……