西北工业大学动力与能源学院上机实习报告(4)

开始 输入年月日 判断是否为闰年 按月选择计算 输出结果 结束

源程序代码：

#include

int ping[] = {0,31,28,31,30,31,30,31,31,30,31,30,31}; int run[] = {0,31,29,31,30,31,30,31,31,30,31,30,31};

bool f(int year) {

if(year % 4 == 0 && year % 100 != 0 || year % 400 == 0) return true; return false; }

int main() {

int year,month,day,i,j;

while(~scanf(\ {

if((year == 0 || month == 0 || day == 0) || (month == 2 && day>29) ||

(run[month]

(!(year%4 == 0 && year0 != 0 ||year@0 == 0)&&month == 2 &&

day == 29) ) {

printf(\ continue; }

if(f(year))

for(j = 0,i = 1;i

for(j = 0,i = 1;i

printf(\ }

return 0; }

程序执行结果(拷屏)：

源程序文件名： 第二天第三题.cpp

6．用最小二乘法求解方程组的近似解

2x+4y =11 3x-5y=3 x+2=6 4x+2y=14

程序编制要点(知识点、程序框图)：

开始 输入矩阵形式的各项系数 运用最小二乘法得到正则方程 用雅可比迭代法解正则方程组 输出方程组的解 结束 源程序代码：

#include #include using namespace std;

int main () {

start:

int m,n,i,j,k;

double sum ,cha,cha1,jing=0.000001; char ch;

cout << \输入方程个数\ cin >> m;

cout << \输入未知数个数\ cin >> n;

double * a = new double [m*n];

double * at = new double [m*n]; double * ata = new double [n*n]; double * b = new double [m]; double * atb = new double [n]; double * x = new double [n];

cout << \请输入化简后系数矩阵，不包含常数项\for (i=0;i> a[i] ;

cout << \请输入化简后右侧常数项\for (i=0;i> b[i];

for (i=0;i

at[j*m+i]=a[i*n+j]; }

for (i=0;i

for (k=0,sum=0;k

sum = sum + at[i*m+k] * a[k*n+j]; ata[i*n+j] = sum; }

for (i=0;i

for (j=0,sum=0;j

sum = sum + at[i*m+j] * b [j]; atb[i] = sum; }

for (i=0;i

x[i]=1; }

for (cha=1;cha>jing;) {

cha1= x[0];

for (i=0;i

for (j=0,sum=0;j

if (i==j) continue;

sum = sum + ata [i*n+j]*x[j]; }

x[i]=(atb[i]-sum)/ata[i*n+i]; }

if (cha1 > x[0]) cha = cha1 -x[0]; else cha = x[0] - cha1; }

if (n==2) {

cout << \ } else {

for (i=0;i

cout << \ } }

delete []a; delete []at; delete []ata; delete []atb; delete []b; delete []x;

cout << \是否继续？y/n\ cin >> ch;

if (ch=='y') goto start; return 0; }

程序执行结果(拷屏)：