1.17 k阶斐波那契数列

已知k 阶斐波那契序列的定义为

；

试编写求k阶斐波那契序列的第m项值的函数算法，k和m均以值调用的形式在函数参数表中出现。

递归算法（较完善，输入引导，反复计算）如下：

#include <stdio.h>
int fibonacci(int layer, int position)
{int fi_value;if (position < layer - 1) {fi_value = 0;return fi_value;}else if (position == layer - 1)return fi_value = 1;else if (position == layer)return fi_value = 1;else{fi_value = 0;fi_value = 2 * fibonacci(layer, position - 1) - fibonacci(layer, position - layer - 1);return fi_value;}
}
void main()
{printf("input layer and position value:(enter other contents to quit)\n");int layer, position,m=10;int ch,scan_return;while ((scan_return=scanf("%d %d", &layer, &position))==2){printf("step1\n");printf("step2\n");printf("layer=%d,position=%d\n", layer, position);if (layer < 0 || position < 0){printf("these two values shouldn't be negative\n\n");continue;}printf("The %dth element of %d layers fibonacci sequence is %d\n", position,layer,fibonacci(layer, position));printf("step3\n\n");printf("input layer and position value:(enter a new line to quit)\n");while ((ch = getchar() != '\n'));}}

1.18 结构处理表

假设有A,B,C,D,E五个高等院校进行田径对抗赛，各院校的单项成绩均已存入计算机，并构成一张表，表中每一行的形式为

编写算法，处理上述表格，以同济各院校的男、女总分和团体总分，并输出。

参考参考答案：

#include <stdio.h>
#define REPORT_NUM 35
typedef enum {A,B,C,D,E}SchoolName;
typedef enum {Male,Female}Gender;
typedef struct {char event[3];Gender sex;SchoolName school;int score;
}Element;typedef struct {int MaleSum;int FemaleSum;int TeamSum;
}Sum;int data_deal(Element*, Sum*);
void main()
{return 1;
}int data_deal(Element* pRecord, Sum* pSum_score)
{pSum_score->MaleSum = 0;pSum_score->FemaleSum = 0;pSum_score->TeamSum = 0;for (int i = 0; i < REPORT_NUM; i++){switch (pRecord[i].sex){case Male:pSum_score[pRecord[i].school].MaleSum += pRecord[i].score;break;case Female:pSum_score[pRecord[i].school].FemaleSum += pRecord[i].score;break;}}for (int i = A; i <= E; i++){pSum_score[i].TeamSum = pSum_score[i].MaleSum + pSum_score[i].FemaleSum;}    return 0;
}

1.19 计算

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>#include <math.h>
#define arrsize 20
unsigned long factorial(int n)
{if (n > 0)return (unsigned long)n * factorial(n - 1);else return 1;
}int compute(unsigned long a[], int n)
{if (n > arrsize){printf("n's too big\n");exit(0);}for (int i = 0;i < n; i++){a[i] = (unsigned long)factorial(i) * pow(2, i);if (a[i] > INT_MAX){printf("result overflow, max_int is %d\n",INT_MAX);exit(OVERFLOW);}}return 1;
}int main()
{int n;unsigned long a[arrsize];printf("enter a n you want limit the computation:\n");while (scanf("%d", &n) == 1){compute(a, n);for (int i = 0; i < n; i++){printf("a[%d]=%d\n", i,a[i]);}printf("enter a n you want limit the computation:\n");}printf("later\n");return 0;
}

1.20 一元多项式

#include <stdio.h>
#include <math.h>
#define arrsize 20
double polydomial(double[], double, int);
int main()
{int n;double a[arrsize], x;printf("enter n:\n");while (scanf("%d", &n) == 1){printf("enter x:\n");if (scanf("%lf", &x) != 1){printf("wrong input!\n");return 0;}for (int i = 0; i <= n; i++){printf("enter a[%d]:\n",i);while (scanf("%lf", &a[i]) != 1){printf("wrong input\n");getchar();}}printf("the value of polydomial is %lf\n", polydomial(a, x, n));printf("enter n:\n");}printf("wrong input\n");
}
double polydomial(double a[], double x, int n)
{double ret_val = 0;for (int i = 0; i <= n; i++){ret_val += a[0] * pow(x, i);}return ret_val;
}