吉大软件专硕967-2006

 

#include
int main(){
    int tra(int n);
    int i,n,add = 0;
    scanf("%d",&n);
    printf("∑di=%d\n",tra(n));
    return 0;
}
int tra(int n){
    int item=n,add=0;
    while(item > 0){
        add += item % 10;
        item /= 10;
    }
    return add;
}

 

 

#include
int main(){
    char **fre(char str[],int n);
    char str[20];
    int i;
    gets(str);
    char **p = fre(str,20);
    for(i=0;*((char *)p+i*2)!='\0';i++){
        printf("字母%c出现的次数:%d\n",*((char *)p+i*2),*((char *)p+i*2+1));
    }
}
char **fre(char str[],int n){
    int i,j,m=0,sign;
    char **fre = (char **)malloc(n*sizeof(char *));
    for(i=0;i<2;i++)
        fre[i] = (char *)malloc(2*sizeof(char));//这两步是申请一个n * 2大小的二维空间
    for(i=0;str[i]!='\0';i++){
        sign = 0;
        if(str[i]>='a' && str[i] <= 'z'){
            for(j=0;j<=m;j++){
                if(*((char *)fre+j*2) == str[i]){
                    *((char *)fre+j*2+1) = *((char *)fre+j*2+1)+1;
                    sign = 1;
                    break;
                }
            }
            if(sign == 0){
                *((char *)fre+m*2) = str[i];
                *((char *)fre+m*2+1) = 1;
                m++;
            }
        }
    }
    *((char *)fre+m*2) = '\0';
    return fre;
}

 

 

3、高精度计算两个长整数的乘积，两个数用数组存放。

#include

#include

#define N 100

#define M 200

int main(){

         intA[N]={0,5,2,3,6,7,3};//3763250

         intB[N]={0,9,0,1};//1090

         intC[N];

         inti,na,nb,n;

         for(na=N-1;na>=0;na--)

                  if(A[na]!=0)break;

         for(nb=N-1;nb>=0;nb--)

                  if(B[nb]!=0)break;

         if(na> nb) n = na;

         elsen = nb;

         for(i=0;i<=n;i++){

                  C[i]=C[i]+ A[i]+B[i];

                  C[i+1]= C[i] / 10;

                  C[i]= C[i] % 10;

         }

         if(C[n+1]!=0)n=n+1;

         printf("restultis:");

         for(i=n;i>=0;i--)

                  printf("%d",C[i]);

        

}

 

 

 

 

 

#include

#include

#define N 100

#define M 200

int main(){

         doublef(double x);

         doublei,j,k,fi,fj,fk;

         i=0;

         j=2;

         fi= f(i);

         fj= f(j);

         while(fabs(fi-fj)> 1e-3){

                  k= (i + j)/2;

                  fk= f(k);

                  if(fk*fi<0){

                          j= k;

                          fj= f(j);

                  }elseif(fk*fj<0){

                          i= k;

                          fi= f(i);

                  }

         }

         printf("f(x)的解为:%.4f\n",i);

        

}

double f(double x){

         return3*x*x*x-5*x-6;

}

 

 

 

 

 

int main(){
	Queue* que = CreateQueue(); //创建一队列 
	struct tree *root,*p;
	int i,j;
	root = creat();
	p = root;
	AddQ(que,p);
	 //每次遍历，找出左右孩子都是叶子的节点，并将此节点变为叶子，值为运算结果
	 //多次遍历，直到这个树只剩下根节点 
	 	printf("%c\n",root->right->symbol);
	while(root->left->symbol!='v' || root->right->symbol!='v'){
		//层次遍历 
		printf("1111");
		while(que!=NULL){
			p = DeleteQ(que);
			if(p->symbol!='v') {
				if((p->left->symbol=='v')&&(p->left->symbol=='v')){
					if(p->symbol=='+')
						p->value = p->left->value + p->right->value;
					else if(p->symbol=='-')
						p->value = p->left->value - p->right->value;
					else if(p->symbol=='*')
						p->value = p->left->value * p->right->value;
					else if(p->symbol=='/')
						p->value = p->left->value / p->right->value;
					p->symbol='v';
			}else{
				if(p->left!=NULL) AddQ(que,p->left);
                if(p->right!=NULL) AddQ(que,p->right);
			}
			}
		}
	}
	printf("%d",root->value);

}

 

 

这道题没太看懂

#include
#include
#define LEN sizeof(struct node)
 struct  node{
	int value1;
	int value2;
	struct node *next;
};
/*创建一个如例图的链表*/
struct node *creat(void){
	struct node *p1,*p2,*p3;
	p1=(struct node *)malloc(LEN);
	p2=(struct node *)malloc(LEN);
	p3=(struct node *)malloc(LEN);
	p1->value1 = 3;
	p1->value2 = 3;
	p2->value1 = -5;
	p2->value2 = 1;
	p3->value1 = 6;
	p3->value2 = 0;
	p1->next=p2;
	p2->next=p3;
	p3->next=NULL;
	return(p1);
}
int main(){
	struct node *p;
	int x,fx=0,n,An;
	scanf("%d",&x);
	p = creat();
	while(p!=NULL){
		An = p->value1;
		n = p->value2;
		fx = fx + An * pow(x,n);
		p=p->next;
	}
	printf("f(x)=%d\n",fx);
}