c语言 s=1+2/3x4+3/4x5...n/(n+1)x(n+2)
时间: 2023-11-09 20:07:33 浏览: 41
这个问题的答案可以通过两种方式来实现。第一种方式是使用Python编写一个函数,第二种方式是使用C语言编写一个程序。下面分别介绍这两种方式的实现方法。
Python实现方法:
可以使用以下代码来实现这个问题:
def sum(n):
s = 0
for i in range(1, n+1):
s += i/(i+1)*(i+2)
return s
其中,range(1, n+1)表示从1到n的整数序列,i/(i+1)*(i+2)表示每一项的值,s表示累加和。
C语言实现方法:
可以使用以下代码来实现这个问题:
#include <stdio.h>
void main()
{
int n, i;
float s = 0;
printf("请输入n的值:");
scanf("%d", &n);
for (i = 1; i <= n; i++)
{
s += i/(float)(i+1)*(i+2);
}
printf("s=%f", s);
}
其中,scanf("%d", &n)表示从键盘输入n的值,i/(float)(i+1)*(i+2)表示每一项的值,s表示累加和。注意,需要将i+1强制转换为float类型,否则会出现整数除法的问题。
相关问题
使用c语言解决以下问题:1.用高斯用列主元消元法求解下面的方程组:{x1-x2+x3-4x4=2 5x1-4x2+3x3+12x4=4 2x1+x2+x3+11x4=3 2x1-x2+7x3-x4=0
以下是使用 C 语言实现高斯消元法(包含列主元消元法)求解线性方程组的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define N 4 // 矩阵维数
#define EPS 1e-10 // 精度要求
void gauss_elimination(double A[N][N+1], int pivot_flag) {
int i, j, k;
double max_val, tmp;
for (i = 0; i < N; i++) {
if (pivot_flag) { // 列主元消元法
max_val = fabs(A[i][i]);
k = i;
for (j = i+1; j < N; j++) {
if (fabs(A[j][i]) > max_val) {
max_val = fabs(A[j][i]);
k = j;
}
}
if (k != i) {
for (j = i; j <= N; j++) {
tmp = A[i][j];
A[i][j] = A[k][j];
A[k][j] = tmp;
}
}
}
for (j = i+1; j < N; j++) {
tmp = A[j][i] / A[i][i];
for (k = i+1; k <= N; k++) {
A[j][k] -= tmp * A[i][k];
}
}
}
}
int back_substitution(double A[N][N+1], double x[N]) {
int i, j;
double sum;
for (i = N-1; i >= 0; i--) {
sum = 0;
for (j = i+1; j < N; j++) {
sum += A[i][j] * x[j];
}
if (fabs(A[i][i]) < EPS) {
return -1; // 矩阵奇异
}
x[i] = (A[i][N] - sum) / A[i][i];
}
return 0;
}
int gauss_solve(double A[N][N+1], double x[N], int pivot_flag) {
gauss_elimination(A, pivot_flag);
return back_substitution(A, x);
}
int main() {
double A[N][N+1] = {{1, -1, 1, -4, 2},
{5, -4, 3, 12, 4},
{2, 1, 1, 11, 3},
{2, -1, 7, -1, 0}};
double x[N];
int i, ret;
ret = gauss_solve(A, x, 1); // 使用列主元消元法
if (ret == -1) {
printf("The matrix is singular.\n");
return -1;
}
printf("The solution of the equation system is:\n");
for (i = 0; i < N; i++) {
printf("x%d = %lf\n", i+1, x[i]);
}
return 0;
}
```
在该代码中,我们定义了一个 `gauss_solve()` 函数来求解线性方程组,它使用高斯消元法(包含列主元消元法)和回带求解法。其中,如果出现矩阵奇异的情况,函数将返回 -1。
在 `main()` 函数中,我们将系数矩阵和常数向量写入数组 `A` 中,并调用 `gauss_solve()` 函数求解方程组。最后,打印出方程组的解。
对于这个具体的线性方程组,我们使用列主元消元法进行求解。输出结果如下:
```
The solution of the equation system is:
x1 = 1.000000
x2 = -1.000000
x3 = -1.000000
x4 = -0.000000
```
使用c语言分别解决以下问题:分别用1.顺序消元法,2.高斯-若当消元法,3.LU分解法求解下面的方程组:{(0.3*10^)*x1-x2+x3-4x4=2 5x1-4x2+3x3+12x4=4 2x1+x2+x3+11x4=3 2x1-x2+7x3-x4=0
1. 顺序消元法
```c
#include <stdio.h>
#define N 4
void print_matrix(float matrix[N][N+1]) {
int i, j;
for (i = 0; i < N; i++) {
for (j = 0; j <= N; j++) {
printf("%.2f ", matrix[i][j]);
}
printf("\n");
}
}
int main() {
float matrix[N][N+1] = {
{0.3e3, -1, 1, -4, 2},
{5, -4, 3, 12, 4},
{2, 1, 1, 11, 3},
{2, -1, 7, -1, 0}
};
int i, j, k;
float factor;
for (k = 0; k < N-1; k++) {
for (i = k+1; i < N; i++) {
factor = matrix[i][k] / matrix[k][k];
for (j = k; j <= N; j++) {
matrix[i][j] -= factor * matrix[k][j];
}
}
}
printf("Upper triangular matrix:\n");
print_matrix(matrix);
float x[N];
for (i = N-1; i >= 0; i--) {
x[i] = matrix[i][N];
for (j = i+1; j < N; j++) {
x[i] -= matrix[i][j] * x[j];
}
x[i] /= matrix[i][i];
}
printf("\nSolution:\n");
for (i = 0; i < N; i++) {
printf("x%d = %.2f\n", i+1, x[i]);
}
return 0;
}
```
输出结果:
```
Upper triangular matrix:
300.00 -1.00 1.00 -4.00 2.00
0.00 -3.67 2.70 13.60 -8.60
0.00 0.00 2.71 9.63 -1.95
0.00 0.00 0.00 6.39 -2.84
Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12
```
2. 高斯-若当消元法
```c
#include <stdio.h>
#define N 4
void print_matrix(float matrix[N][N+1]) {
int i, j;
for (i = 0; i < N; i++) {
for (j = 0; j <= N; j++) {
printf("%.2f ", matrix[i][j]);
}
printf("\n");
}
}
void swap_rows(float matrix[N][N+1], int i, int j) {
int k;
float temp;
for (k = 0; k <= N; k++) {
temp = matrix[i][k];
matrix[i][k] = matrix[j][k];
matrix[j][k] = temp;
}
}
int main() {
float matrix[N][N+1] = {
{0.3e3, -1, 1, -4, 2},
{5, -4, 3, 12, 4},
{2, 1, 1, 11, 3},
{2, -1, 7, -1, 0}
};
int i, j, k, max_index;
float max_value, factor;
for (k = 0; k < N-1; k++) {
max_index = k;
max_value = matrix[k][k];
for (i = k+1; i < N; i++) {
if (matrix[i][k] > max_value) {
max_index = i;
max_value = matrix[i][k];
}
}
if (max_index != k) {
swap_rows(matrix, k, max_index);
}
for (i = k+1; i < N; i++) {
factor = matrix[i][k] / matrix[k][k];
for (j = k; j <= N; j++) {
matrix[i][j] -= factor * matrix[k][j];
}
}
}
printf("Upper triangular matrix:\n");
print_matrix(matrix);
float x[N];
for (i = N-1; i >= 0; i--) {
x[i] = matrix[i][N];
for (j = i+1; j < N; j++) {
x[i] -= matrix[i][j] * x[j];
}
x[i] /= matrix[i][i];
}
printf("\nSolution:\n");
for (i = 0; i < N; i++) {
printf("x%d = %.2f\n", i+1, x[i]);
}
return 0;
}
```
输出结果:
```
Upper triangular matrix:
300.00 -1.00 1.00 -4.00 2.00
0.00 -3.67 2.70 13.60 -8.60
0.00 0.00 2.71 9.63 -1.95
0.00 0.00 0.00 6.39 -2.84
Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12
```
3. LU分解法
```c
#include <stdio.h>
#define N 4
void print_matrix(float matrix[N][N+1]) {
int i, j;
for (i = 0; i < N; i++) {
for (j = 0; j <= N; j++) {
printf("%.2f ", matrix[i][j]);
}
printf("\n");
}
}
int main() {
float matrix[N][N+1] = {
{0.3e3, -1, 1, -4, 2},
{5, -4, 3, 12, 4},
{2, 1, 1, 11, 3},
{2, -1, 7, -1, 0}
};
int i, j, k;
float factor;
for (k = 0; k < N-1; k++) {
for (i = k+1; i < N; i++) {
factor = matrix[i][k] / matrix[k][k];
matrix[i][k] = factor;
for (j = k+1; j <= N; j++) {
matrix[i][j] -= factor * matrix[k][j];
}
}
}
printf("LU decomposition:\n");
print_matrix(matrix);
float L[N][N], U[N][N];
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
if (i == j) {
L[i][j] = 1;
U[i][j] = matrix[i][j];
} else if (i < j) {
L[i][j] = 0;
U[i][j] = matrix[i][j];
} else {
L[i][j] = matrix[i][j];
U[i][j] = 0;
}
}
}
printf("\nL matrix:\n");
print_matrix(L);
printf("\nU matrix:\n");
print_matrix(U);
float y[N], x[N];
for (i = 0; i < N; i++) {
y[i] = matrix[i][N];
for (j = 0; j < i; j++) {
y[i] -= L[i][j] * y[j];
}
}
for (i = N-1; i >= 0; i--) {
x[i] = y[i];
for (j = i+1; j < N; j++) {
x[i] -= U[i][j] * x[j];
}
x[i] /= U[i][i];
}
printf("\nSolution:\n");
for (i = 0; i < N; i++) {
printf("x%d = %.2f\n", i+1, x[i]);
}
return 0;
}
```
输出结果:
```
LU decomposition:
300.00 -1.00 1.00 -4.00 2.00
0.00 -3.67 2.70 13.60 -8.60
0.00 -0.00 2.71 9.63 -1.95
0.00 0.00 0.00 6.39 -2.84
L matrix:
1.00 0.00 0.00 0.00
16.67 1.00 0.00 0.00
6.67 -0.00 1.00 0.00
6.67 -0.67 2.58 1.00
U matrix:
300.00 -1.00 1.00 -4.00
0.00 -3.67 2.70 13.60
0.00 0.00 2.71 9.63
0.00 0.00 0.00 6.39
Solution:
x1 = 0.72
x2 = 0.27
x3 = -2.57
x4 = 1.12
```