第28次CCF计算机软件能力认证(测试)

测试300分要是考试的时候也能这么发挥就好

第一题：现值计算

解题思路：直接模拟

n , m = input().split()
n = int(n);m = float(m)
l = list(map(int , input().split()))
res = 0
for i in range(0 , n + 1):res += pow(1 + m , -i) * l[i]
print(res)

第二题：训练计划

解题思路：读题+模拟

#include<iostream>
#include<cstring>using namespace std;const int N = 400;
int p[N] , t[N];
int n , m;int main()
{cin >> n >> m;for(int i = 1;i <= m;i ++) cin >> p[i];for(int i = 1;i <= m;i ++) cin >> t[i];int first_time[N];bool f = true;for(int i = 1;i <= m;i ++){if(!p[i]) first_time[i] = 1;else first_time[i] = t[p[i]] + first_time[p[i]];int check = first_time[i] + t[i] - 1;if(check > n) f = false;}for(int i = 1;i <= m;i ++)cout << first_time[i] << " ";if(f){cout << endl;memset(first_time , 0x3f , sizeof first_time);for(int i = m;i >= 1;i --){if(first_time[i] > n) first_time[i] = n + 1 - t[i];if(p[i]) first_time[p[i]] = min(first_time[p[i]] , first_time[i] - t[p[i]]);}for(int i = 1;i <= m;i ++)cout << first_time[i] << " ";cout << endl;}return 0;
}

第三题：JPEG编码

 解题思路：依旧是模拟，但是更大

# z型遍历
import math
def cal(i , j , m):res = 0for u in range(8):for v in range(8):alphau = 1 if u else (0.5) ** 0.5alphav = 1 if v else (0.5) ** 0.5ucos = math.cos(math.pi * (1 / 8) * (i + (1 / 2)) * u)vcos = math.cos(math.pi * (1 / 8) * (j + (1 / 2)) * v)res += 0.25 * (alphau * alphav * m[u][v] * ucos * vcos)return resdef input_m(data):M = [[0 for i in range(8)] for j in range(8)]x = y = 0f = Truewhile x != 8 or y != 8:if x < 8 and y < 8:if not len(data):breakM[x][y] = data[0]data.pop(0)if f: x -= 1;y += 1else: x += 1;y -= 1if x < 0: x = 0;f = not fif y < 0: y = 0;f = not freturn Mdef cal_muti(Q , M):res = [[0 for i in range(8)] for j in range(8)]for i in range(8):for j in range(8):res[i][j] = Q[i][j] * M[i][j]return resQ = [] # 量化矩阵
for i in range(8):l = list(map(int , input().split()))Q.append(l)
n = int(input())
t = int(input()) # t = 0 输出填充矩阵
data = list(map(int , input().split()))
input_M = input_m(data)
liang_M = cal_muti(Q , input_M)
if t == 0 or t == 1:for i in range(8):for j in range(8):print(input_M[i][j] if t == 0 else liang_M[i][j] , end = ' ')print()
else:for i in range(8):for j in range(8):res = round(cal(i , j , liang_M) + 128)if res < 0: res = 0if res > 255: res = 255print(res, end = ' ')print()