problem 1178

   稍微推一下公式就可以了。
Accepted1178C00:00.01392K
#include<stdio.h>
void solve(int  n)
{
    
int s = n / 4 + 1
;
    
if (n % 4 == 0) s--
;
    
int a = s * 4
;
    
int
 i;
    
for (i = 1; i <= s; i++
)
    {
        
int x = a - 2 * i + 2
;
        
int y = 2 * i - 1
;
        
if (!(x > n && y >
 n))
        {
            printf(
"Sheet %d, front: "
,i);
            
if (x > n)    printf("Blank, "
);
            
else printf("%d, "
,x);
            
if (y > n)    printf("Blank "
);
            
else    printf("%d/n"
,y);
        }

        x 
= 2 *
 i;
        y 
= a - 2 * i + 1
;
        
if (!(x > n && y >
 n))
        {
            printf(
"Sheet %d, back : "
,i);
            
if (x > n)    printf("Blank, "
);
            
else printf("%d, "
,x);
            
if (y > n)    printf("Blank/n"
);
            
else    printf("%d/n"
,y);
        }
    }
}
void
 main()
{
#ifndef ONLINE_JUDGE
    freopen(
"1178.txt","r"
,stdin);
#endif

    
int  n;
    
while(scanf("%d",&n) != EOF && n != 0
)
    {
        printf(
"Printing order for %d pages:/n"
,n);
        solve(n);
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
#endif

}
 
service->sin_addr.s_addr = inet_addr(hostname); if (service->sin_addr.s_addr == INADDR_NONE) { struct hostent *host = gethostbyname(hostname); if (host == NULL || host->h_addr == NULL) { RTMP_Log(RTMP_LOGERROR, "Problem accessing the DNS. (addr: %s)", hostname); ret = FALSE; goto finish; } service->sin_addr = *(struct in_addr *)host->h_addr; } 上面是代码的一部分,现在出现这个报错, INFO: ....RTMP chn [0] CONNECT........... ERROR: RTMP chn [0] ONLINE, try to connect. connecttimes: 4 [Error]2025-10-13/14:46:25-567 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 [cloud-brd](tid3069517824)1760337987192|2305|4|cloud_ip_list.c:56:cloud_ip_list_parse| - get addr info error, host:n-device-sur.tplinkcloud.com, serv:443, Name or service not known [cloud-brd](tid3069517824)1760337987192|2305|4|cloud_session.c:1515:cloud_session_connect_defaultSvr| - fail to get ip [cloud-brd](tid3069517824)1760337987192|2305|4|cloud_session.c:665:cloud_session_active_long_time| - fail to connect the defaultSvr, result:-4 [Error]2025-10-13/14:46:30-568 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 [Error]2025-10-13/14:46:35-572 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 [Error]2025-10-13/14:46:40-572 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 [Error]2025-10-13/14:46:45-573 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 [Error]2025-10-13/14:46:50-577 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 ERROR: Problem accessing the DNS. (addr: ns8.indexforce.com) ERROR: RTMP chn [0] enter recon_protection mode. [Error]2025-10-13/14:46:55-578 preview_simple_add_batch_auth: Unauth Channel: 0 / 2. Try next time. PreviewPage.c:1178 INFO: ....RTMP chn [0] CONNECT........... 而且这两次INFO: ....RTMP chn [0] CONNECT...........之间花了30s,你帮我看看这上面的代码哪里导致阻塞的
最新发布
10-14
Dynamic programming is an algorithm suitable for solving the Knapsack problem. It avoids repeated calculations by decomposing complex problems into overlapping sub - problems and storing the solutions to the sub - problems. The core of dynamic programming is state definition and state transition equations [^1]. The general steps of using dynamic programming to solve the Knapsack problem are as follows: 1. **Define the state**: Usually, two - dimensional states are defined. For example, let `dp[i][j]` represent the maximum value that can be obtained when considering the first `i` items and the capacity of the knapsack is `j`. 2. **State transition equation**: For the 0 - 1 Knapsack problem, if the weight of the `i` - th item is `w[i]` and the value is `v[i]`, then the state transition equation is: - When `j < w[i]`, `dp[i][j]=dp[i - 1][j]` (the current item cannot be put into the knapsack). - When `j >= w[i]`, `dp[i][j]=max(dp[i - 1][j], dp[i - 1][j - w[i]]+v[i])` (choose whether to put the current item into the knapsack). Here is a simple Python code example for the 0 - 1 Knapsack problem: ```python def knapsack(weights, values, capacity): n = len(weights) dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)] for i in range(1, n + 1): for j in range(1, capacity + 1): if j < weights[i - 1]: dp[i][j] = dp[i - 1][j] else: dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weights[i - 1]] + values[i - 1]) return dp[n][capacity] weights = [2, 3, 4, 5] values = [3, 4, 5, 6] capacity = 8 print(knapsack(weights, values, capacity)) ``` ### Application scenarios - **Resource allocation**: In project management, given a limited amount of resources (such as time, budget, manpower), and different tasks with different resource requirements and benefits, the Knapsack problem can be used to select the most profitable combination of tasks. - **Stock selection**: When an investor has a certain amount of funds and there are multiple stocks with different prices and expected returns, the Knapsack problem can be used to select the most profitable stock portfolio.
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