长宁区公司网站建设/网络舆情分析师

eg2.写一个函数printN，打印出从1到N

#include<stdio.h>void printN1(int n);  //直接输出 
void printN2(int n);  //递归 void printN2(int n){if(n){printN2(n-1);printf("%-4d ", n);		}
}void printN1(int n){int i;for(i = 0; i < n; i++){printf("%-4d ", i+1);}
}void main(void){int n;printf("请输入一个正整数：");scanf("%d", &n);printN1(n);printN2(n);
}

eg3.写程序计算给定多项式在给定点x处的值

#include<stdio.h>
#include<math.h>#define  N    3double fun1(double a[], int n, double x);  //a放系数，n是次幂 
double fun2(double a[], int n, double x); double fun2(double a[], int n, double x){double y = a[n];int i; for(i = n; i > 0; i--){y = a[i-1]+ y*x;//printf("y=%lf\n", y);		}return y;
}double fun1(double a[], int n, double x){double y = 0;int i; for(i = 0; i <= n; i++){y += a[i]*pow(x, i);  }return y;
}void main(void){double a[N+1];int i;double x;printf("请输入多项式的系数(次幂逐渐增大,最高次幂为%d)：", N);for(i = 0; i < N+1; i++){scanf("%lf", &a[i]);}printf("请输入x的值：");scanf("%lf", &x);printf("fun1函数在点%lf处的值为：%lf\n", x, fun1(a, N, x));printf("fun2函数在点%lf处的值为：%lf\n", x, fun2(a, N, x));} 

最大子列和问题(遗留，分而治之)

//最大子列和问题
//8
//-2 6 -1 5 4 -7 2 3
//结果：14 #include<stdio.h>#define  N   8void inputData(int a[], int n);
int maxSubseqSum1(int a[], int n);   //T(n)=O(n^3)  把所有子序列的和都求出来 
int maxSubseqSum2(int a[], int n);   //T(n)=O(n^2)  把所有子序列的和都求出来 
int maxSubseqSum2(int a[], int n);   //T(n)=O(nlogn)  分而治之，不会 int maxSubseqSum2(int a[], int n){int i;int j;int k; int sum;int max = 0;for(i = 0; i < n; i++){sum = 0;for(j = i; j < n; j++){sum+=a[j];if(sum > max){max = sum;}} }return max;}int maxSubseqSum1(int a[], int n){int i;int j;int k; int sum;int max = 0;for(i = 0; i < n; i++){for(j = i; j < n; j++){sum = 0;for(k = i; k < j; k++){sum+=a[k];}if(sum > max){max = sum;}} }return max;
}void inputData(int a[], int n){int i;printf("请输入%d个整型数据：", N);for(i = 0; i < n; i++){scanf("%d", &a[i]);}
}void main(void){int a[N];inputData(a, N);printf("最大子列和为：%d\n", maxSubseqSum1(a, N));printf("最大子列和为：%d\n", maxSubseqSum2(a, N));
}