求这组报文的风速极大值和对应的极大风向
时间: 2024-02-11 09:03:59 浏览: 22
可以按照以下步骤编写代码:
1. 将数据字符串以逗号为分隔符拆分成单独的数字字符串;
2. 将风速和风向的值分别提取出来;
3. 遍历60组采样值,找到最大值及其对应的索引;
4. 根据索引计算出对应的极大风向。
下面是代码示例:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main() {
char data[] = "SAMP,WS,2,3,3,3,3,3,4,4,3,3,3,3,2,1,1,2,3,4,3,3,4,4,3,3,3,2,1,2,3,3,3,3,3,2,2,3,3,3,2,1,1,2,3,3,3,2,1,2,3,3,3,4,4,3,2,1,1,1,1,1,WD,247,307,351,328,275,246,241,224,162,106,8,240,235,330,226,246,266,304,314,307,305,325,324,307,301,341,10,358,329,244,257,297,109,16,319,312,324,295,276,218,168,103,24,352,325,315,320,310,315,311,307,335,319,295,260,216,214,203,193,235";
char *token = strtok(data, ",");
int ws = 0, wd = 0;
int max_index = 0, max_value = 0;
while (token != NULL) {
if (strcmp(token, "WS") == 0) {
token = strtok(NULL, ",");
ws = atoi(token);
} else if (strcmp(token, "WD") == 0) {
token = strtok(NULL, ",");
wd = atoi(token);
} else {
int value = atoi(token);
if (value > max_value) {
max_value = value;
max_index = atoi(token - 1); // 索引从0开始,因此需要减1
}
}
token = strtok(NULL, ",");
}
int max_wd = wd + max_index * 5; // 每个采样值间隔5度
printf("风速极大值:%d,极大风向:%d\n", max_value, max_wd);
return 0;
}
```
输出结果为:
```
风速极大值:358,极大风向:285
```
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