fortran programming language tutorial

Problem DescriptionThis day the teacher again to small Hao out a topic: give you three lengths, respectively, A,b,c match, let you calculate the three matches can make up of the Triangle area.InputEnter each row to include three number a,b,c.OutputThe area of the output triangle for each row (reserved three decimal places), cannot be composed of triangles told the small Hao output "no!".Sample Input1 2 33) 4 5Sample Outputno!6.000HINT1#include 2#include 3 intMain ()4 {5 Doubles,a,b,c,p;6

Problem DescriptionThe teacher gave a title to Xiao Hao: give you a value of n (within 32 bits) in seconds, let you convert to: minutes: seconds output. Now that little Howe is in a quandary, please help him solve the problem.InputEnter a positive integer of n seconds per line.Outputfor each set of test data, output a row of results: minutes: seconds. (24-hour system)Sample Input560Sample Output00:00:0500:01:00HINT#include intHh,mm,ss;voidConvertime (Longsec) {hh=sec/3600; MM= (sec-hh*3600L)/ -;

The title describes the sum of the diagonal elements of a 3x3 matrix. Input matrix output main diagonal sub-diagonal element and sample input1 2 31 1 13 2 1Sample output3 71#include 2 3 intMainintargcChar Const*argv[])4 {5 inta[3][3], I, J;6 intDiagonal1 =0, Diagonal2 =0;7 8 //input9 for(i =0; I 3; i++)Ten { One for(j =0; J 3; J + +) A { -scanf"%d", a[i][j]); - } the } - -Diagonal1 = a[0][0] + a[1][1] +a[2][2]; -Diagonal2 = a[2][0] + a[1][1] +a[

The title description writes a function that enables a given two-dimensional array (3x3) to transpose, that is, the row and column interchange. Enter a 3x3 matrix output to transpose matrices sample input1 2 34 5 67 8 9Sample output1#include 2 3 4 voidConvertinta[][3],intRowintCol//two-dimensional array when making formal parameters, the second must be written5 {6 intI, J, temp;7 for(i =0; i )8 {9 for(j = i; J )Ten { One //Exchange A if(I! =j

].birthday.month, S[i].birthday.day, - s[i].department, s[i].major); WuPutchar ('\ n'); - } About } $ - voidSearchInfo (Studentinfo s[], date date) - { - inti; A for(i =0; i ){ + if(s[i].birthday.month>date.month) { theprintf"\n%4d%16s%2d/%2d", S[i].num,s[i].name, S[i].birthday.month, - s[i].birthday.day); $ Continue; the } the if(S[i].birthday.month = = Date.month s[i].birthday.day>date.day) { theprintf"\n%4d%16s%2d/%2d", S[i].num, S[i].name, S

parameter. Call to function The basic format of the function call statement is: The The parameters in the Example: 1 intdataarray[ -];2 intN;3 DoubleAve;4 5 //Custom Functions6 DoubleAverageintValue[],intnum) {7 intI, sum =0;8 for(i =0; i ){9Sum + =Value[i];Ten returnsum*1.0/num; One } A } - - //function Call theAve = Average (DataArray, n);Application examples of custom functions Output multiplication Ta

Title DescriptionFor the sum of the following three digits, the sum of 2 decimal 1~a and the reciprocal of the square and 1~c of the 1~b are reserved.InputA b COutput1+2+...+a + 1^2+2^2+...+b^2 + 1/1+1/2+...+1/cSample input100 50 10Sample output47977.93TipsNote the output format, do not output extra space!Source#includemain(){int i,a,b,c;float s=0;scanf("%d%d%d",a,b,c);for(i=1;is+=i;for(i=1;is+=i*i;for(i=1;is+=1.0/i;printf("%.2f\n",s);}1016:c Language

Problem DescriptionThis day the teacher gave small Hao a very simple topic relaxed: Enter a score, let you find their simplest score. InputThe first line includes a T, which represents the number of groups of test data.The next line of T lines includes a fraction. (The numerator denominator is within the int range)OutputFor each test sample, the output line has its simplest fraction.Sample Input32/63/111/36Sample Output1/33/111/36My Code:1#include 2ProcessintAintb/*greatest common divisor, so th

Problem DescriptionBeg sn=2+22+222+...+22 ... A value of 222 (with N 2). For example: 2+22+222+2222+22222 (n=5), n is entered by the keyboard.InputNOutputAndSample Input5Sample Output24690My Code:1#include 2 voidMain ()3 {4 intN,a=2, t,sum=0, I;5scanf"%d",n);6t=A;7 for(i=1; i)8 {9sum=sum+T;Tent=t*Ten+A; One } Aprintf"%d", sum); -}Other code:1#include 2 3 using namespacestd;4 5 intMain ()6 {7 Longsum=0;8 intt=2;9 intN;TenCin>>N; One for(intI=0; ii) A { -Sum + =T;

Problem DescriptionThe teacher gave a title to Xiao Hao: give you two integers x and n ( -10now that little Howe is in a quandary, please help him solve the problem. InputEnter two integers x and n per line.Outputfor each set of test data, the output line evaluates. Sample Input1 12 2Sample Output16HINT1#include 2 3 4 Long LongSumintXintN)5 {6 Long Longs=0;7 inti,j;8 Long LongS1;9 for(i=1; i)Ten { Ones1=1; A for(j=1; j) - { -s1*=x; the } -s+=S1; - } -