c语言：输入一个五位整数将数中的五个数值进行从大到小的顺序排序形成一个新的五位整数并输出这个整数。 c语言：输入一个五位数，对此整数中的五个数值进行从大到小排序...
\u8f93\u5165\u4e00\u4e2a\u4e94\u4f4d\u6574\u6570\uff0c\u5bf9\u6b64\u6574\u6570\u4e2d\u7684\u4e94\u4e2a\u6570\u503c\u8fdb\u884c\u4ece\u5927\u5230\u5c0f\u7684\u987a\u5e8f\u6392\u5e8f\uff0c\u5f62\u6210\u4e00\u4e2a\u65b0\u7684\u4e94\u4f4d\u6574\u6570\uff0c\u5e76\u8f93\u51fa\u8fd9\u4e2a\u6574\u6570\u4f7f\u7528\u5192\u6ce1\u6392\u5e8f
void bubble(int *a,int n) /*\u5b9a\u4e49\u4e24\u4e2a\u53c2\u6570\uff1a\u6570\u7ec4\u9996\u5730\u5740\u4e0e\u6570\u7ec4\u5927\u5c0f*/ { int i,j,temp; for(i=0;ia[j]) { temp=a[i]; a[i]=a[j]; a[j]=temp; } \uff5d\uff5d}
/*
\u7f16\u5199\u4e00\u4e2a\u7a0b\u5e8f\uff0c\u8ba9\u5b83\u6709\u4ee5\u4e0b\u529f\u80fd\uff1a\u4ece\u952e\u76d8\u4e0a\u8f93\u5165\u4e00\u4e2a\u4e94\u4f4d\u6570\uff0c\u5bf9\u6b64\u6574\u6570\u4e2d\u7684\u4e94\u4e2a\u6570\u503c
\u8fdb\u884c\u4ece\u5927\u5230\u5c0f\u6392\u5e8f\uff0c\u5f62\u6210\u4e00\u4e2a\u65b0\u7684\u4e94\u4f4d\u6570\uff0c\u8f93\u51fa\u8fd9\u4e2a\u6574\u6570\u3002\uff0815\u5206\uff09
*/
#include
#include
void main()
{
long x;
/*\u7528\u6570\u7ec4a\u5b58\u50a8\u4e94\u4f4d\u6570\u7684\u4e94\u4f4d*/
/*x1,x2,x3,\u5206\u522b\u8868\u793a\u4e94\u4f4d\u6570\u53d6\u4f5910000,1000,100\u540e\u7684\u6570*/
long a[5],x1,x2,x3;
int i,j,temp=0;
printf("input a number:");
scanf("%ld",&x); //\u5bf9\u5e94\u8d77\u6765\u3002
a[0]=x/10000;
x1=x%10000;
a[1]=x1/1000;
x2=x1%1000;
a[2]=x2/100;
x3=x2%100;
a[3]=x3/10;
a[4]=x%10;
for(i=0;i<5-1;i++) //n\u4e2a\u6570 \u6392n-1\u8d9f
for(j=0;j<5-i-1;j++) //\u8fd9\u91cc\u4e0d\u662f 5-i \u82e5i=0\uff0c\u5219j+1 \u8d8a\u754c\u4e86\u3002
{
if(a[j]<a[j+1])
{
temp=a[j];
a[j]=a[j+1];
a[j+1]=temp;
}
}
printf("\nnew number is:%ld",a[0]*10000+a[1]*1000+a[2]*100+a[3]*10+a[4]);
getch();
}
根据题意得到如下代码：
#include <stdio.h>
int getMaxFive(int n)
{
int a[5], i, j, k = 0, t;
while (n!=0){
a[k] = n%10;
n /= 10;
k++;
}
if (k != 5)return -1;
for (i = 0; i < k; ++i){
for (j = 0; j < k-i-1; ++j){
if (a[j] < a[j+1]){
t = a[j], a[j] = a[j+1], a[j+1] = t;
}
}
}
t = 0;
for (i = 0; i < k; ++i){
t = t*10 + a[i];
}
return t;
}
int main()
{
int e = getMaxFive(82731);
printf ("%d
", e);
return 0;
}
#include <stdio.h>
struct student
{
char name[15];
struct student *next;
};
struct student *link(struct student *a, struct student *b) {
struct student *p = a;
while (p->next)
p = p->next;
p->next = b; // p->next = b->next; 这里你确定链表是有头指针还是没有头指针吗？
return a;
}
int main() {
struct student stu1, stu2;
stu1.next = NULL;
stu2.next = NULL;
scanf("%s %s", stu1.name, stu2.name);
printf("%s", link(&stu1, &stu2));
}
绛旓細include<stdio.h> main(){ int a,b,c,d,e,n;printf("璇杈撳叆涓涓簲浣嶆暟锛");scanf("%d",&n);a=n/10000;n=n%10000;b=n/1000;n=n%1000;c=n/100;n=n%100;d=n/10;n=n%10;e=n;printf("浜斾釜浣嶆暟涔嬪拰涓猴細%d\n",a+b+c+d+e);} 杩欎釜绛旀澶熺畝鍗曚簡鍚э紝涓鐩簡鐒讹紒锛侊紒
绛旓細浣犵殑c璇█绋嬪簭鎴戝府浣犳敼濂戒簡(瑙佸浘,鏀瑰姩鐨勫湴鏂硅娉ㄩ噴,浠呬袱澶)灏辨槸scanf鍑芥暟涓殑鍙橀噺a[i]鍓嶉潰娌″姞鍙栧湴鍧绗
绛旓細p++;}/* 绋嬪簭鐨勮緭鍑洪『搴忎负 10 -> 8 -> 6 -> 4 -> 2 閫掑綊姝ラ瑙ｉ噴 1: 绗竴娆℃墽琛宻er鍑芥暟,numbser = 12345鍒ゆ柇鏁板瓧澶т簬9,灏唅nt(1234)鍐嶆鎵ц鍑芥暟 ps: (int)/(int) 缁撴灉涓篿nt 2: 寰呮墽琛宻er鍑芥暟鍚,杈撳嚭缁撴灉涓綅鏁 12345 % 10 = 5 3: 绗簩娆℃墽琛宻er鍑芥暟...绫绘瘮姝ラ1 4: 寰...
绛旓細鏂规硶寰堝鍢 闅忎究鏉涓涓細include <stdio.h> include <stdlib.h> void main(){ char p[10];char *q;int i=0;scanf("%s",p);q=p;while(*(q++))i++;q--;while(i-->=0)printf("%c",*(q--));}
绛旓細include<stdio.h> void main(){ int n,i,a[10];printf("璇杈撳叆涓涓涓嶈秴杩5浣嶇殑姝鏁存暟锛\n");scanf("%d",&n);int sum=0;for(i=0;;i++){ a[i]=n%10;/*鍙栦綑瀛樹簬a[i]涓*/ n=n/10;sum++;/*浣嶆暟鍔犱竴*/ if(n==0) break;} printf("瀹冩槸%d浣嶆暟\n",sum);int j;printf...
绛旓細///鍔熻兘锛5浣嶆暣鏁板弽杞 ///浣滆咃細鎮夐洦杈板瘋 /// //澶存枃浠 include <stdio.h> include <string.h> //涓诲嚱鏁 main(){ char str_in[6];char str_out[6];printf("璇杈撳叆涓涓簲浣嶆暣鏁帮細")scanf("%s",str_in);str_out=my_strrev(str_in);printf("鍙嶈浆鍚庣殑浜斾綅鏁存暟鏄細%s",str_out);}...
绛旓細include <stdio.h>int main(void){ int a[5],i; printf("杈撳叆5涓暣鏁帮細"); for(i=0;i<5;i++) scanf("%d",&a[i]); printf("灏5涓暣鏁板悇鑷姞1鍚庝负锛"); for(i=0;i<5;i++) printf("%d ",a[i]+1); printf("\n"); return 0;} ...
绛旓細(1)include<stdio.h> void main(){ char str[5];int i;gets(str);for(i=4;i>=0;i--)printf("%c",str[i]);} (2)include<stdio.h> void main(){ int n,a,b,c,d,e,m;scanf("%d",&n);a=n%10000;b=(n-a*10000)%1000;c=(n-a*10000-b*1000)%100;d=(n-a*10000...
绛旓細;//杈撳嚭浣嶆暟 for(;size > 0;size--) { n *= 10; printf("%d",(int)n % 10);//姝ｅ悜杈撳嚭姣忎綅鐨鏁板瓧 } return 0;}浣跨敤娴偣鏁颁笉濂斤紝鎴戞祴璇200浠ュ唴鐨勬暟锛杈撳叆34銆68銆170銆178銆188锛岀粨鏋滄瘮鏁板瓧灏忎竴锛屽叾浠栫殑姝ｅ父锛屾渶濂借繕鏄娇鐢ㄦ暟缁勫惂銆
绛旓細include <stdio.h>void fun(int n){int m=0;while(n){m=m*10+n%10;n/=10;}while(m){printf("%d ",m%10);m/=10;}}void main(){int n;scanf("%d",&n);fun(n);}杩愯绀轰緥锛
