用c语言如何在96个float类型的数据中找到众数（相关搜索：函数的调用|数组长度）

给定两组等长的数据，用函数求两组数据的众数，所有变量全用 float 型；（main 函数中只能有函数的调用， 不允许存在任何计算过程） 两组数据如下↓，每个数组长度为 96；
float Temp_X[96] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.46667, 11.4667, 31.6, 52.7333, 80.3333, 116.333, 156.6, 199.4, 242.2, 283.4, 329.2, 379.333, 431.333, 482.6, 541, 594.4, 643.533, 692.133, 736.267, 772.667, 810.133, 841.867, 868.2, 892.4, 917.667, 939.8, 954.667, 969, 976.8, 983.4, 987.467, 994.933, 1023.67, 875.2, 873.933, 758.8, 678.2 , 515.867, 782.533, 908.8, 779.2, 831.4, 645.533, 734.067, 679.533, 610.267, 565.067, 512.467 , 462, 405.2, 354.133, 302, 247.8, 191.533, 140, 94.2667, 57.5333, 25.9333, 4, 0, 0, 0, 0, 0, 0, 0 , 0, 0, 0, 0 }; // x
float Temp_Y[96] = { 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 1.785, 2.57833, 3.927, 5.79233, 7.379, 9.48133, 11.1473, 12.4167, 13.6627, 16.193701, 18.248699, 19.042, 19.042, 19.105301, 16.6383, 17.240999, 14.631, 11.8217, 11.663, 12.155, 15.488, 21.859301, 19.32, 19.042, 19.6133, 21.105, 22.9937, 20.827299, 23.858299, 23.0333, 19.2883, 15.6937, 21.5893, 23.802999, 20.518299, 21.5893, 17.907301, 17.971001, 17.574301, 16.781, 15.5513, 12.3773, 10.2747, 8.60867, 6.86333, 5.39567, 3.88767, 2.856, 2.142, 2.142, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952 }; //y

引用 皆我百晓生 小程序回复内容作答：
要在96个float类型的数据中找到众数，可以按以下步骤操作：

1. 首先，定义一个函数float mode(float arr[], int len)，该函数接受一个float类型的数组和数组的长度作为输入参数，并返回一个float类型的值作为众数。
2. 在mode函数中，定义一个变量maxCount，用于保存当前出现最多的次数，初始值设为0。定义一个变量modeValue，用于保存众数的值，初始值设为0。
3. 使用for循环遍历数组中的每个元素。
   • 在循环内部，定义一个变量count，用于记录当前元素出现的次数，初始值设为0。
   • 使用另一个for循环，再次遍历数组中的每个元素，统计与当前元素相同的个数。
   • 如果count大于maxCount，就将maxCount更新为count，并将modeValue更新为当前元素的值。
4. 循环结束后，返回modeValue作为众数的值。
5. 在主函数main中，调用mode函数分别传入Temp_X和Temp_Y数组作为参数，将返回的众数值打印出来。

以下是完整的代码示例：

#include <stdio.h>

float mode(float arr[], int len) {
    float modeValue = 0;
    int maxCount = 0;

    for (int i = 0; i < len; i++) {
        int count = 0;
        for (int j = 0; j < len; j++) {
            if (arr[j] == arr[i]) {
                count++;
            }
        }
        if (count > maxCount) {
            maxCount = count;
            modeValue = arr[i];
        }
    }

    return modeValue;
}

int main() {
    float Temp_X[96] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.46667, 11.4667, 31.6, 52.7333, 80.3333, 116.333, 156.6, 199.4, 242.2, 283.4, 329.2, 379.333, 431.333, 482.6, 541, 594.4, 643.533, 692.133, 736.267, 772.667, 810.133, 841.867, 868.2, 892.4, 917.667, 939.8, 954.667, 969, 976.8, 983.4, 987.467, 994.933, 1023.67, 875.2, 873.933, 758.8, 678.2 , 515.867, 782.533, 908.8, 779.2, 831.4, 645.533, 734.067, 679.533, 610.267, 565.067, 512.467 , 462, 405.2, 354.133, 302, 247.8, 191.533, 140, 94.2667, 57.5333, 25.9333, 4, 0, 0, 0, 0, 0, 0, 0 , 0, 0, 0, 0 };
    float Temp_Y[96] = { 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 1.785, 2.57833, 3.927, 5.79233, 7.379, 9.48133, 11.1473, 12.4167, 13.6627, 16.193701, 18.248699, 19.042, 19.042, 19.105301, 16.6383, 17.240999, 14.631, 11.8217, 11.663, 12.155, 15.488, 21.859301, 19.32, 19.042, 19.6133, 21.105, 22.9937, 20.827299, 23.858299, 23.0333, 19.2883, 15.6937, 21.5893, 23.802999, 20.518299, 21.5893, 17.907301, 17.971001, 17.574301, 16.781, 15.5513, 12.3773, 10.2747, 8.60867, 6.86333, 5.39567, 3.88767, 2.856, 2.142, 2.142, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952 };

    float mode_x = mode(Temp_X, 96);
    float mode_y = mode(Temp_Y, 96);

    printf("众数 x: %f\n", mode_x);
    printf("众数 y: %f\n", mode_y);

    return 0;
}

【以下回答由 GPT 生成】

#include <stdio.h>

float findMode(float arr[], int n) {
    float mode = arr[0];
    int count = 1;
    int maxCount = 1;

    for(int i = 1; i < n; i++) {
        if(arr[i] == arr[i-1]) { // 如果当前元素和前一个元素相同
            count++; // 计数器加1
        } else { // 如果当前元素和前一个元素不同
            count = 1; // 重置计数器
        }

        if(count > maxCount) { // 如果当前计数器值大于最大计数器值
            maxCount = count; // 更新最大计数器值
            mode = arr[i]; // 更新众数
        }
    }

    return mode;
}

int main() {
    // 调用findMode函数
    float Temp_X[96] = { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1.46667, 11.4667, 31.6, 52.7333, 80.3333, 116.333, 156.6, 199.4, 242.2, 283.4, 329.2, 379.333, 431.333, 482.6, 541, 594.4, 643.533, 692.133, 736.267, 772.667, 810.133, 841.867, 868.2, 892.4, 917.667, 939.8, 954.667, 969, 976.8, 983.4, 987.467, 994.933, 1023.67, 875.2, 873.933, 758.8, 678.2 , 515.867, 782.533, 908.8, 779.2, 831.4, 645.533, 734.067, 679.533, 610.267, 565.067, 512.467 , 462, 405.2, 354.133, 302, 247.8, 191.533, 140, 94.2667, 57.5333, 25.9333, 4, 0, 0, 0, 0, 0, 0, 0 , 0, 0, 0, 0 };

    float Temp_Y[96] = { 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 0.595, 1.785, 2.57833, 3.927, 5.79233, 7.379, 9.48133, 11.1473, 12.4167, 13.6627, 16.193701, 18.248699, 19.042, 19.042, 19.105301, 16.6383, 17.240999, 14.631, 11.8217, 11.663, 12.155, 15.488, 21.859301, 19.32, 19.042, 19.6133, 21.105, 22.9937, 20.827299, 23.858299, 23.0333, 19.2883, 15.6937, 21.5893, 23.802999, 20.518299, 21.5893, 17.907301, 17.971001, 17.574301, 16.781, 15.5513, 12.3773, 10.2747, 8.60867, 6.86333, 5.39567, 3.88767, 2.856, 2.142, 2.142, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952, 0.952 };

    float mode_X = findMode(Temp_X, 96);
    float mode_Y = findMode(Temp_Y, 96);

    printf("Temp_X的众数为: %f\n", mode_X);
    printf("Temp_Y的众数为: %f\n", mode_Y);

    return 0;
}

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