General test and Machine Test

I. Auto-increment and auto-increment operations

++ I: I increases by 1 and then participates in other operations.
-- I: I minus 1 and then participates in other operations.
I ++: After I is involved in the operation, the value of I increases by 1.
I --: After I is involved in the operation, the value of I is reduced by 1.

Ii. integer-based floating point number in C Language

Two methods:

1. Float F = .....;
Int I = (INT) (F + 0.5 );
I is the result of rounding F.

2. Use library functions

Floor: round down
Floor (2.6) = 2
Floor (-2.6) =-3
Ceil: rounded up
Ceil (2.3) = 3
Ceil (-2.3) =-3
Be sure to include the header file math. h.
References:
1. http://topic.csdn.net/t/20050203/14/3773229.html
2. http://zhidao.baidu.com/question/155536957.html
3. http://www.360doc.com/content/11/0429/17/3931678_113204596.shtml

III. C language variable Memory Allocation

References:

1. http://blog.csdn.net/subo86/article/details/4814874

2. http://topic.csdn.net/t/20051110/09/4383464.html

3. http://zhidao.baidu.com/question/52843150.html

Iv. number of replies

1. determine whether an integer is a return number:

# Include <stdio. h> void main () {int I, num, new_num = 0, temp = 0; printf ("Enter the number to be judged:"); scanf ("% d ", & num); temp = num; // convert the number of inputs into the reverse! = 0) {int gewei = TEMP % 10; new_num = new_num * 10 + gewei; temp = temp/10 ;} // If a number is the same as the original number, if (num = new_num) {printf ("% d", num ); printf ("this number is the number of input \ n");} else {printf ("% d", num); printf ("this number is not the number of input \ n ");}}

References: http://topic.csdn.net/u/20080331/22/d36bf390-944e-432b-a31b-d186dd4db4d0.html

2. Determine whether a string is a return number:

# Include <stdio. h ># include <string> void daozhi (char STR []) {int I, n; char C, * s = STR; For (n = 0; s [N]! = '\ 0';) n ++; printf ("n = % d \ n", n); for (I = 0; I <= n/2; I ++) {c = STR [I]; STR [I] = STR [n-i-1]; STR [n-i-1] = C ;}} void main () {char str1 [100], str2 [100]; int A; gets (str1); strcpy (str2, str1); daozhi (str2); A = strcmp (str1, str2); if (a = 0) printf ("% s is input \ n", str1); else printf ("% s is not input \ n", str1 );}

References: http://zhidao.baidu.com/question/123856313.html

V. Reverse sorting of linked lists

# Include <iostream> using namespace STD; typedef struct node {int data; node * Next;} * lklist; // generate the linked list lklist creatlist (int n) {lklist head by means of end insertion, p, q; int size = sizeof (node); Head = (lklist) malloc (size); P = head; int X; For (INT I = 0; I <N; I ++) {q = (lklist) malloc (size); cout <"input node" <I <"value:"; CIN> X; q-> DATA = x; P-> next = Q; P = Q;} p-> next = NULL; return head ;} // output a linked list in reverse order void printlk (node * l) {lklist p, q; P = q = L;/* P, Q is the two pointers pointing to the header Node */While (p-> next! = NULL) P = p-> next;/* point P to the last node of the key table */while (1) {While (Q-> next! = P) q = Q-> next;/* Ask Q to find the last node to be printed */printf ("% d \ n ", p-> data); P = Q;/* P move one forward */q = L;/* q points to the header node */If (P = L) /* exit after access */break;} // exchange the lklist reverse (lklist h) before and after the element position of the linked list // H is the head pointer of the linked list {lklist P, V1, v2; v2 = H; V1 = NULL; while (V2! = NULL) {P = v2-> next; v2-> next = V1; V1 = V2; v2 = P;} return V1 ;} // swap the positions of elements in the linked list before and after lklist res (lklist h) {lklist S, S1; S = H; H = NULL; while (s) {S1 = s; S = s-> next; S1-> next = H; H = S1;} return h;} void main () {lklist LK = creatlist (5 ); // printlk (lk); // lklist rk = reverse (lk); lklist rk = res (lk); While (rk-> next! = NULL) {cout <rk-> data <"; rk = rk-> next ;}}

References:

1. http://zhidao.baidu.com/question/41056423.html
2. http://lixon2000.blog.163.com/blog/static/996498020061026113420769/

6. Big integer multiplication

void multiply(char* a,char* b,char* c){    int i,j,ca,cb,* s;    ca=strlen(a);    cb=strlen(b);    s=(int*)malloc(sizeof(int)*(ca+cb));    for (i=0;i<ca+cb;i++)        s[i]=0;    for (i=0;i<ca;i++)        for (j=0;j<cb;j++)            s[i+j+1]+=(a[i]-'0')*(b[j]-'0');    for (i=ca+cb-1;i>=0;i--)        if (s[i]>=10)        {            s[i-1]+=s[i]/10;            s[i]%=10;        }    i=0;    while (s[i]==0)        i++;       for (j=0;i<ca+cb;i++,j++)           c[j]=s[i]+'0';    c[j]='\0';    free(s);}

References:

1. http://topic.csdn.net/u/20080203/10/64d519c1-030c-4725-abe4-e4d1061e894c.html
2. http://wenku.baidu.com/view/8ac7c44ffe4733687e21aadd.html
3. http://wenku.baidu.com/view/968c4ba0b0717fd5360cdc71.html

VII. Newton Iteration Method for Finding the root of the equation

#include<stdio.h>float solution(float x){float x1,y,k;do{k=6*x*x-8*x+3;y=2*x*x*x-4*x*x+3*x-6;x1=x-y/k;x=x1;}while(fabs(y)<0.001);return x;}void main(){float x;x=1.5;x=solution(x);printf("%f\n",x);}

References:

1. http://zhidao.baidu.com/question/73178644.html? Fr = qrl & cid = 866 & Index = 4

2. http://zhidao.baidu.com/question/6326922.html? Fr = qrl & cid = 93 & Index = 5 & FR2 = Query

 

It is a classic program to determine whether an integer is a program that multiply the number of input and the number of integers. Sometimes it is difficult to think.