数组分段求和 - 编程论坛

#2

不会游泳的虾2021-09-15 11:19

供参考：

程序代码：

//假设数组a有1152个数，我想按照顺序每100个数算一个和，求一个平均值，该如何实现？

//假设数组a有1152个数，我想按照顺序每100个数算一个和，求一个平均值，该如何实现？

#include<stdio.h>
#include<time.h>
#include <stdlib.h>
int main()
{
    int a[1152], i, k, c, s[13] = { 0 };
    srand((unsigned int)time(NULL));
    for (i = 0, k = 0, c = 0; i < 1152; i++) {
        a[i] = rand() % 150 + 1;
        s[k] += a[i];
        c++;
        if (c == 100) {
            k++;
            c = 0;
        }
    }
    for (i = 0; i < k; i++){
        printf(" %c %d",i==0?'(':'+', s[i]);
        s[12] += s[i];
    }
    printf(" ) / %d = %d \n",k, s[12] / k);
    return 0;
}

#3

自由而无用2021-09-15 17:16

//online parser: https://www.bccn.net/run/

程序代码：

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[])
{
    char c[12][100];
    int i, j, sum;

    memcpy(c, main, 1152);

    for(i = 0; i < 12; i++) {
        for(j = 0, sum = 0; j < sizeof(c[i]); j++) sum += c[i][j];
        printf("%02d: sum = %d, avr = %d\n", i + 1, sum, sum / 100);
    }

    return 0;
}

[此贴子已经被作者于2021-9-15 17:34编辑过]

#4

我善治鬼2021-09-17 14:13

程序代码：

#include <stdio.h>

#define N 1152

int main()
{
    int sum = 0;
    for (int i = 0, j; i < N; i++) {
        sum += i;
        j = (i + 1) % 100;
        if (j == 0 || i == N - 1) {
            if (j == 0) j = 100;
            printf("%d / %d = %d\n\n", i, j, sum / j);
            sum = 0;
        }
        else printf("%d + ", i);
    }
    return 0;
}

#5

纯蓝之刃2021-09-19 20:55

程序代码：

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main()
{
    unsigned int num_all,num_one,num_current,group,i,j;
    float *a,sum;
    printf("请输入总个数及每组个数:");
    scanf("%u %u",&num_all,&num_one);
    a=malloc(num_all*sizeof(float));

    srand((unsigned int)time(NULL));    //模拟数据
    for(i=0; i<num_all; i++)
        *(a+i)=rand();

    group=num_all/num_one;
    if(num_all%num_one!=0)
        group+=1;

    for(j=0; j<group; j++)
    {
        sum=0;
        if(j+1==group)
        {
            if(num_all%num_one!=0)
                num_current=num_all%num_one;
        }
        else
        {
            num_current=num_one;
        }

        for(i=0; i<num_current; i++)
        {
            sum+=*(a+j*num_one+i);
        }
        printf("第%u组和为%.2f,平均值为%.2f\n",j,sum,sum/num_current);
    }

    free(a);
    return 0;
}