離散數(shù)學(xué)逆函數(shù)c語(yǔ)言代碼,離散數(shù)學(xué)與c程序設(shè)計(jì)

C語(yǔ)言寫(xiě)函數(shù)fun求一個(gè)整數(shù)的逆序數(shù)，在main中輸入兩個(gè)整數(shù)，求其逆序數(shù)之和并輸出。如輸入32 -71 輸出6

#include stdio.h

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int fun(int num);

void main( )

{

int x,y,sum;

sum=0;

printf("請(qǐng)輸入兩個(gè)整數(shù)：");

scanf("%d%d",x,y);

sum=fun(x)+fun(y); %因?yàn)閒un（）函數(shù)有返回值，這里相當(dāng)于將兩個(gè)返回值相加

printf("sum=%d\n",sum);

}

int fun(int num)

{

int a,b; %定義 a，b變量

a=0;

while (num!=0) %判斷循環(huán)的結(jié)束條件 此處num以兩位數(shù)為例

{

b=num%10; %對(duì)10取余，即求出num的個(gè)位數(shù)賦值給b

a=a*10+b;

num=num/10; %將num的末位數(shù)去掉

}

return a;

}

C語(yǔ)言編程：編寫(xiě)一個(gè)函數(shù)求逆矩陣

#include?stdio.h

#include?stdlib.h

#include?malloc.h

void?MatrixOpp(double?*A,?int?m,?int?n,?double*?invmat);

void?MatrixInver(double?*A,?int?m,?int?n,?double*?invmat);

double?Surplus(double?A[],?int?m,?int?n);

int?matrix_inv(double*?p,?int?num,?double*?invmat);

void?MatrixOpp(double?A[],?int?m,?int?n,?double*?invmat)

{

int?i,?j,?x,?y,?k;

double?*SP?=?NULL,?*AB?=?NULL,?*B?=?NULL,?X;

SP?=?(double?*)?malloc(m?*?n?*?sizeof(double));

AB?=?(double?*)?malloc(m?*?n?*?sizeof(double));

B?=?(double?*)?malloc(m?*?n?*?sizeof(double));

X?=?Surplus(A,?m,?n);

X?=?1?/?X;

for?(i?=?0;?i??m;?i++)

for?(j?=?0;?j??n;?j++)

{

for?(k?=?0;?k??m?*?n;?k++)

B[k]?=?A[k];

{

for?(x?=?0;?x??n;?x++)

B[i?*?n?+?x]?=?0;

for?(y?=?0;?y??m;?y++)

B[m?*?y?+?j]?=?0;

B[i?*?n?+?j]?=?1;

SP[i?*?n?+?j]?=?Surplus(B,?m,?n);

AB[i?*?n?+?j]?=?X?*?SP[i?*?n?+?j];

}

}

MatrixInver(AB,?m,?n,?invmat);

free(SP);

free(AB);

free(B);

}

void?MatrixInver(double?A[],?int?m,?int?n,?double*?invmat)

{

int?i,?j;

double?*B?=?invmat;

for?(i?=?0;?i??n;?i++)

for?(j?=?0;?j??m;?j++)

B[i?*?m?+?j]?=?A[j?*?n?+?i];

}

double?Surplus(double?A[],?int?m,?int?n)

{

int?i,?j,?k,?p,?r;

double?X,?temp?=?1,?temp1?=?1,?s?=?0,?s1?=?0;

if?(n?==?2)

{

for?(i?=?0;?i??m;?i++)

for?(j?=?0;?j??n;?j++)

if?((i?+?j)?%?2)

temp1?*=?A[i?*?n?+?j];

else

temp?*=?A[i?*?n?+?j];

X?=?temp?-?temp1;

}

else

{

for?(k?=?0;?k??n;?k++)

{

for?(i?=?0,?j?=?k;?i??m,?j??n;?i++,?j++)

temp?*=?A[i?*?n?+?j];

if?(m?-?i)

{

for?(p?=?m?-?i,?r?=?m?-?1;?p??0;?p--,?r--)

temp?*=?A[r?*?n?+?p?-?1];

}

s?+=?temp;

temp?=?1;

}

for?(k?=?n?-?1;?k?=?0;?k--)

{

for?(i?=?0,?j?=?k;?i??m,?j?=?0;?i++,?j--)

temp1?*=?A[i?*?n?+?j];

if?(m?-?i)

{

for?(p?=?m?-?1,?r?=?i;?r??m;?p--,?r++)

temp1?*=?A[r?*?n?+?p];

}

s1?+=?temp1;

temp1?=?1;

}

X?=?s?-?s1;

}

return?X;

}

int?matrix_inv(double*?p,?int?num,?double*?invmat)

{

if?(p?==?NULL?||?invmat?==?NULL)

{

return?1;

}

if?(num??10)

{

return?2;

}

MatrixOpp(p,?num,?num,?invmat);

return?0;

}

int?main()

{

int?i,?j;

int?num;

double?*arr=NULL;

double?*result=NULL;

int?flag;

printf("請(qǐng)輸入矩陣維數(shù)：\n");

scanf("%d",num);

arr=(double?*)malloc(sizeof(double)*num*num);

result=(double?*)malloc(sizeof(double)*num*num);

printf("請(qǐng)輸入%d*%d矩陣：\n",?num,?num);

for?(i?=?0;?i??num;?i++)

{

for?(j?=?0;?j??num;?j++)

{

scanf("%lf",?arr[i?*?num?+?j]);

}

}

flag?=?matrix_inv(arr,?num,?result);

if(flag==0)

{

printf("逆矩陣為：\n");

for?(i?=?0;?i??num?*?num;?i++)

{

printf("%lf\t?",?*(result?+?i));

if?(i?%?num?==?(num?-?1))

printf("\n");

}

}

else?if(flag==1)

{

printf("p/q為空\(chéng)n");

}

else

{

printf("超過(guò)最大維數(shù)\n");

}

system("PAUSE");

free(arr);

free(result);

return?0;

}

c語(yǔ)言 離散數(shù)學(xué)集合復(fù)合運(yùn)算的代碼，（R。R）的代碼實(shí)現(xiàn)

//說(shuō)明：輸入的格式需要提示按輸入，因?yàn)橐@取正確的有序?qū)Σ拍苓M(jìn)行復(fù)合運(yùn)算

/*

*************輸入格式如:a b, #,# 退出***************

輸入:a b

輸入:b t

輸入:t d

輸入:s j

輸入:j i

輸入:c a

*/

#include "stdlib.h"

typedef char Element;

struct Node

{

Element left;

Element right;

struct Node *next;

};

struct Node *CreateLink();

struct Node *Operation(struct Node *R,struct Node *S);

void PrintLink(struct Node *h);

int main()

{

struct Node *hdR,*hdS,*rhd;

printf("請(qǐng)輸入第一個(gè)集合R的關(guān)系\n");

hdR = CreateLink();

PrintLink(hdR);

printf("\n請(qǐng)輸入第二個(gè)集合S的關(guān)系\n");

hdS = CreateLink();

PrintLink(hdS);

rhd = Operation(hdR,hdS);

if (rhd-next == NULL)

{

printf("\nR。S結(jié)果為空集\n");

}

else

{

printf("\nR。S結(jié)果為:\n");

PrintLink(rhd);

}

return 0;

}

struct Node *CreateLink()

{

struct Node *head, *p;

printf("*************輸入格式如:a b, \'#,#\' 退出***************\n");

Element a,b;

head = (struct Node *)malloc(sizeof(struct Node));

head-left = 0;

head-right = 0;

head-next = NULL;

printf("輸入:");

scanf("%c %c",a,b);

getchar();

while (a != '#')

{

p = (struct Node *)malloc(sizeof(struct Node));

p-left = a;

p-right = b;

p-next = head-next;

head-next = p;

printf("輸入:");

scanf("%c %c",a,b);

getchar();

}

return head;

}

struct Node *Operation(struct Node *R, struct Node *S)

{

struct Node *newHead,*newP,*newQ;

struct Node *pH, *pNext;

newHead = (struct Node *)malloc(sizeof(struct Node));

newHead-left = 0;

newHead-right = 0;

newHead-next = NULL;

newP = newHead;

if (R-next == NULL || S-next == NULL)

{

return newP;

}

char cLeft,cRight;

pH = R;

while (pH-next != NULL)

{

cLeft = pH-next-left;

cRight = pH-next-right;

pNext = S-next;

while(pNext != NULL)

{

//存在可以復(fù)合運(yùn)算的

if (cRight == pNext-left)

{

//在復(fù)合運(yùn)算結(jié)果集中插入數(shù)據(jù)，如果存在相同的二元關(guān)系，則不需要插入

newP = newHead;

while (newP-next != NULL)

{

if (cLeft == newP-left cRight == newP-right)

{

break;

}

newP = newP-next;

}

if (newP-next == NULL)

{

newQ = (struct Node *)malloc(sizeof(struct Node));

newQ-left = cLeft;

newQ-right = pNext-right;

newQ-next = NULL;

newP-next = newQ;

// newQ-next = newP-next-next;

}

}

pNext = pNext-next;

}

pH = pH-next;

}

return newHead;

}

void PrintLink(struct Node *h)

{

struct Node *p=h-next;

printf("\n");

while (p != NULL)

{

printf("%c,%c ",p-left,p-right);

p = p-next;

}

printf("\n");

}

c語(yǔ)言求逆序數(shù)的代碼調(diào)試

#includestdio.h

#includemath.h

int?reserve(int?number){

int?i=1,sum=0,n;

while(number/(int)pow(10,i)0){//?這里用(int)pow(10,i)取整數(shù)，因?yàn)閜ow函數(shù)返回的是浮點(diǎn)型值

++i;}

while(i!=0){

n=number%10;

number=number/10;

sum=sum+n*pow(10,i-1);

i=i-1;

}

return?sum;

}

int?main(void){

int?num,re;

printf("請(qǐng)輸入一個(gè)整數(shù):");

scanf("%d",num);

re=reserve(num);

printf("逆序數(shù):%d\n",re);

return?0;

}

參考修改過(guò)的代碼哈，歡迎交流，滿意請(qǐng)采納。

在C語(yǔ)言中如何定義逆序數(shù)函數(shù)

思路是先轉(zhuǎn)成字符串再操作

返回值為計(jì)算出的逆序數(shù)

int f(int num)

{

char an[15];

int i, len, t, neg = 0;

if(num 0)

{

num = -num;

neg = 1;

}

sprintf(an, "%d", num);

len = strlen(an);

for(i=0; ilen/2; i++)

{

t = an[i];

an[i] = an[len - 1 -i];

an[len - 1 -i] = t;

}

num = atoi(an);

return (neg?-num:num);

}

剛才沒(méi)看到你還沒(méi)學(xué)到字符串，再給你個(gè)簡(jiǎn)單點(diǎn)的

int f(int num)

{

int a=0,b;

while (num != 0)

{

b=num%10;

a=a*10+b;

num=num/10;

}

return a;

}

給定一個(gè)從{1,2,...,n}到其自身的函數(shù)f，判斷函數(shù)f是否有反函數(shù)？ 怎么用c語(yǔ)言編程解決?。?/h2>

首先是要知道理論, 有反函數(shù)需要的是雙射, 對(duì)于有限集到自身的映射來(lái)說(shuō), 單射和滿射等價(jià), 所以這里只需要判斷是否是滿射就可以了.

然后是技術(shù)上的實(shí)現(xiàn). 如果有O(n)的存貯空間(比如開(kāi)設(shè)一個(gè)長(zhǎng)度為n的數(shù)組a[], 初始化成0), 那么遍歷一遍 k = 1,...,n, 置a[f(k)+1] = 1, 最后遍歷一遍 a 就可以判定 f 是否是滿射, 時(shí)間復(fù)雜度是O(n).

(如果實(shí)在沒(méi)有辦法開(kāi)設(shè)額外的存貯空間, 那么至少可以按定義去判定 f 是否是單射, 時(shí)間復(fù)雜度是O(n^2).)

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