#define M 3 #define N M+1 #define NN N*N/2 int main( ){ printf( "%d," , NN ) ; printf( "%d" , 5 * NN ) ; return 0 ; }
时间: 2023-06-17 08:04:11 浏览: 77
程序输出的结果是:6,30。
在程序中,首先定义了M为3,然后定义N为M+1,即N为4。接着定义NN为N*N/2,即NN为2*2*4/2,即NN为4。
在程序中,第一个printf语句输出的是NN的值,即6。第二个printf语句输出的是5*NN的值,即5*6,即30。因此程序输出的结果为6,30。
相关问题
#include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <nanomsg/nn.h> #include <nanomsg/reqrep.h> #define SERVER_ADDRESS "tcp://127.0.0.1:5555" // 服务器地址 int main() { int sock = nn_socket(AF_SP, NN_REQ); // 创建一个REQ类型的socket if (sock < 0) { fprintf(stderr, "nn_socket error: %s\n", nn_strerror(nn_errno())); return -1; } if (nn_connect(sock, SERVER_ADDRESS) < 0) { // 连接服务器地址 fprintf(stderr, "nn_connect error: %s\n", nn_strerror(nn_errno())); nn_close(sock); return -1; } char *request = "123"; int request_len = strlen(request) + 1; int bytes = nn_send(sock, request, request_len, 0); // 发送请求消息 if (bytes < 0) { fprintf(stderr, "nn_send error: %s\n", nn_strerror(nn_errno())); nn_close(sock); return -1; } printf("client sent: %s\n", request); char *buf = NULL; bytes = nn_recv(sock, &buf, NN_MSG, 0); // 接收回复消息 if (bytes < 0) { fprintf(stderr, "nn_recv error: %s\n", nn_strerror(nn_errno())); nn_close(sock); return -1; } printf("client received: %s\n", buf); nn_freemsg(buf); // 释放接收到的消息内存 nn_close(sock); // 关闭socket return 0; } 以上是一个客户端代码,客户端发送内容“123”服务器回复“abc”,我想要更改一下,客户端发送一段json,使用libjansson,json内容为{"module":"1","from":"2","time":"","service":"get_dp_version","args":[]}
你可以将原来的请求消息改成一个json字符串,然后使用libjansson库将其序列化成一个字符数组,再发送给服务器。修改后的代码如下所示:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <nanomsg/nn.h>
#include <nanomsg/reqrep.h>
#include <jansson.h>
#define SERVER_ADDRESS "tcp://127.0.0.1:5555" // 服务器地址
int main() {
int sock = nn_socket(AF_SP, NN_REQ); // 创建一个REQ类型的socket
if (sock < 0) {
fprintf(stderr, "nn_socket error: %s\n", nn_strerror(nn_errno()));
return -1;
}
if (nn_connect(sock, SERVER_ADDRESS) < 0) { // 连接服务器地址
fprintf(stderr, "nn_connect error: %s\n", nn_strerror(nn_errno()));
nn_close(sock);
return -1;
}
json_t *root = json_object(); // 创建一个json对象
json_object_set_new(root, "module", json_string("1")); // 设置module字段
json_object_set_new(root, "from", json_string("2")); // 设置from字段
json_object_set_new(root, "time", json_string("")); // 设置time字段
json_object_set_new(root, "service", json_string("get_dp_version")); // 设置service字段
json_object_set_new(root, "args", json_array()); // 设置args字段为空数组
char *request = json_dumps(root, 0); // 将json对象序列化成字符串
int request_len = strlen(request) + 1;
int bytes = nn_send(sock, request, request_len, 0); // 发送请求消息
if (bytes < 0) {
fprintf(stderr, "nn_send error: %s\n", nn_strerror(nn_errno()));
nn_close(sock);
return -1;
}
printf("client sent: %s\n", request);
char *buf = NULL;
bytes = nn_recv(sock, &buf, NN_MSG, 0); // 接收回复消息
if (bytes < 0) {
fprintf(stderr, "nn_recv error: %s\n", nn_strerror(nn_errno()));
nn_close(sock);
return -1;
}
printf("client received: %s\n", buf);
nn_freemsg(buf); // 释放接收到的消息内存
nn_close(sock); // 关闭socket
free(request); // 释放序列化后的字符串内存
json_decref(root); // 释放json对象内存
return 0;
}
```
请用C语言实现Once upon a time in the mystical land of Draconis, there existed two powerful arrays: M and N . These arrays were filled with positive integers, each carrying its own magical essence. The inhabitants of the land were intrigued by the concept of similarity between arrays. They discovered that two arrays, M and N , could be considered similar if it was possible to transform a subarray of N into M by adding or subtracting a constant value to each element. You are now summoned to solve a puzzle. Given two arrays, M and N , your task is to determine the number of subarrays of N that are similar to M . Will you be able to unravel this mystical connection? Input The input consists of multiple lines. The first line contains two integers M and N (1≤M≤N≤106) , representing the lengths of arrays M and N respectively. The second line contains M space-separated positive integers m1,m2,…,mM (1≤mi≤109) , representing the magical elements of array M . The third line contains N space-separated positive integers n1,n2,…,nN (1≤ni≤109) , representing the mystical elements of array N . Output Output a single integer, the number of subarrays of N that are similar to M .
为了解决这个问题,我们可以先对两个数组进行排序,然后用双指针法来找到所有相似的子数组。具体实现如下:
```c
#include <stdio.h>
#include <stdlib.h>
#define MAX_N 1000000
int M[MAX_N], N[MAX_N];
int m, n;
int compare(const void *a, const void *b) {
return *(int*)a - *(int*)b;
}
int count_similar() {
int i = 0, j = 0;
int count = 0;
while (i < m && j < n) {
if (M[i] == N[j]) { // 找到相同的元素
int k = j;
while (k < n && N[k] == N[j]) {
k++;
}
count += k - j; // 加上与N[j]相同的元素的个数
i++;
j = k;
} else if (M[i] < N[j]) { // M[i]较小,需要增加N中的元素
i++;
} else { // M[i]较大,需要减少N中的元素
j++;
}
}
return count;
}
int main() {
scanf("%d%d", &m, &n);
for (int i = 0; i < m; i++) {
scanf("%d", &M[i]);
}
for (int i = 0; i < n; i++) {
scanf("%d", &N[i]);
}
qsort(M, m, sizeof(int), compare);
qsort(N, n, sizeof(int), compare);
printf("%d\n", count_similar());
return 0;
}
```
首先读入两个数组的长度和元素,然后对它们进行排序。然后用双指针法找到所有相似的子数组,具体方法如下:
1. 初始化两个指针i和j,分别指向M和N的开头;
2. 如果M[i]等于N[j],说明找到了一个相似的子数组,接着向右移动j,直到N[j]不等于N[j+1];
3. 如果M[i]小于N[j],说明N[j]需要增加,所以向右移动i;
4. 如果M[i]大于N[j],说明N[j]需要减少,所以向右移动j;
5. 重复2~4步,直到i到达M的末尾或j到达N的末尾。
遍历过程中,我们用一个计数器count来累计相似的子数组的个数,每当找到一个相似的子数组时,我们就把与N[j]相同的元素的个数加到count中。最后返回count即可。
时间复杂度为O(MlogM + NlogN),空间复杂度为O(M + N)。
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