Question

C语言问题，求大神解答？

Answer 1

#include<stdio.h>

void main()

{

char ch;

int a;

ch=getchar();

if(ch>='a'&&ch<='z')

ch='1';

if(ch>='A'&&ch<='Z')

ch='2';

if(ch==' ')

ch='3';

switch(ch)

{

case'1':

printf("小写字母\n");

break;

case'2':

printf("大写字母\n");

break;

case'3':

printf("空格\n");

break;

default:

printf("其他");

}

}

Answer 2

#include "stdafx.h"
#include <stdio.h>

//定义矩阵的维度
#define ROWCNT 3 //行数量
#define COLCNT 4 //列数量

//加数1
int V1[]={15,22,-6,9};
int B1[ROWCNT*COLCNT]={1,0,0,1,0,1,0,0,1,0,0,0};
//加数2
int V2[]={1,2,3,4,5};
int B2[ROWCNT*COLCNT]={1,1,0,1,0,1,0,0,1,0,0,0};
//和
int V[ROWCNT*COLCNT]={0,0,0,0,0,0,0,0,0,0,0,0};
int B[ROWCNT*COLCNT]={0,0,0,0,0,0,0,0,0,0,0,0};

void MatrixAdd(int *aV,int *aB,int *bV,int *bB,int *sumV,int *sumB){
int i;
for(i=0;i<ROWCNT*COLCNT;i++){
if ((*aB==1)&&(*bB==0)){
*sumV=(*aV);
*sumB=1;
aV++;
sumV++;
}
else if ((*aB==0)&&(*bB==1)){
*sumV=(*bV);
*sumB=1;
bV++;
sumV++;
}
else if((*aB==1)||(*bB==1)){
*sumV=(*aV)+(*bV);
*sumB=1;
aV++;
bV++;
sumV++;
}
else{
*sumB=0;
}
aB++;
bB++;
sumB++;
}
}

void Show(int *aV,int *aB){
int i,j,k=0;
for(i=0;i<ROWCNT;i++){
for(j=0;j<COLCNT;j++){
if(*aB==0){
printf("%d ",*aB);
}
else{
printf("%d ",*aV);
aV++;
}
aB++;
}
printf("\n");
}
}

int main(int argc, char* argv[]){
int i=0;
printf("-第一个矩阵---\n");
Show(V1,B1);
printf("-第二个矩阵---------\n");
Show(V2,B2);

MatrixAdd(V1,B1,V2,B2,V,B);//求和
printf("-和---------\n");
Show(V,B);
scanf("%d",i);
return(0);
}
因为计算累加的时间与矩阵的数量是线性关系，所以复杂度为O(n)