求简单C语言程序代码!
输入2个正整数m和n,求其最大公约数和最小公倍数#include#includeint main()int m,n,p,q,s,r;printf("请输入两个正整数;m,nn");scanf("%d,%d",&m,&n);#includemain()int a,b,t=0;scanf("%d %d",&a,&b);if (a小游戏2048源码:#include#include#include#include#includeint jsk( ); //计算空格数void rsgm( ); //重置游戏void inkey( ); //按键输入void left( ); //向左移动void right( ); //向右移动void up( ); //向上移动void down( ); //向下移动void show( ); //输出界面void adnum( ); //添加随机数void yes( ); //游戏是否结束(1是0否)void gtxy(int x, int y); //控制光标位置的函数int a[4][4]; //存储16个格子中的数字int score = 0; //每局得分int best = 0; //最高得分int ifnum; //是否需要添加数字(1是0否)int over; //游戏结束标志(1是0否)int i,j,k;int main( ){ rsgm( );//重置游戏inkey( );//按键输入return 0;}void setColor(unsigned short ForeColor = 7, unsigned short BackGroundColor = 0){ HANDLE handle = GetStdHandle(STD_OUTPUT_HANDLE);SetConsoleTextAttribute(handle, ForeColor + BackGroundColor * 0x10);} //用于控制字符颜色的函数void rsgm( )//重置游戏{ score = 0; ifnum = 1; over = 0; srand((unsigned)time(0)); //启动随机数发生器int n = rand( ) % 16; //随机函数产生0-15的数字for (i = 0; i < 4; i++){for (j = 0; j < 4; j++){ if (n == 0) { int k = rand( ) % 3; if (k == 0 || k == 1) { a[i][j] = 2; }else { a[i][j] = 4; } n--; }else { a[i][j] = 0; n--; }}}adnum( );system("cls");CONSOLE_CURSOR_INFO cursor_info={1,0}; //以下两行是隐藏光标的设置SetConsoleCursorInfo(GetStdHandle(STD_OUTPUT_HANDLE),&cursor_info);setColor(14, 0); //设置字体淡红色,背景为黑色printf("nntt 2048小游戏"); setColor(7, 0); //恢复白字黑底printf("nt┌──────┬──────┬──────┬──────┐");printf("nt│ │ │ │ │");printf("nt├──────┼──────┼──────┼──────┤");printf("nt│ │ │ │ │");printf("nt├──────┼──────┼──────┼──────┤");printf("nt│ │ │ │ │");printf("nt├──────┼──────┼──────┼──────┤");printf("nt│ │ │ │ │");printf("nt└──────┴──────┴──────┴──────┘");show( );}void show( )//输出界面{ for(i=0;i<4;i++)for(j=0;j<4;j++){ gtxy(7*j+9,2*i+4); //gtxy(7*j+9, 2*i+4)是光标到指定位置输出数字if(a[i][j]==0){printf(" "); setColor(7, 0); printf("│");}else if(a[i][j]<10){ if (a[i][j] == 2) {setColor(14, 0); }else if (a[i][j] == 4) {setColor(13, 0); }else if (a[i][j] == 8) {setColor(12, 0); }printf(" %d ", a[i][j]); setColor(7, 0); printf("│");}else if (a[i][j] < 100){if (a[i][j] == 16) {setColor(12, 0); }else if (a[i][j] == 32) {setColor(10, 0); }else if (a[i][j] == 64) {setColor(2, 0); }printf(" %d ", a[i][j]); setColor(7, 0); printf("│");}else if (a[i][j] < 1000) {if (a[i][j] == 128) {setColor(9, 0); }else if (a[i][j] == 256) {setColor(1, 0); }else if (a[i][j] == 512) {setColor(13, 0); }printf(" %d ", a[i][j]); setColor(7, 0); printf("│");}else if (a[i][j] < 10000) {if (a[i][j] == 1024) {setColor(5, 0); }else {setColor(15, 0); }printf(" %d ", a[i][j]); setColor(7, 0); printf("│");}}if (jsk( ) == 0){ yes( ); if (over) { gtxy(9,12); setColor(10, 0);printf("nn 游戏结束!是否继续? [ Y/N ]:"); }}}void inkey( )//按键输入{ int key;while (1){ key = getch( );if (over) { if (key == 89|| key == 121) {rsgm( ); continue; }else if (key == 78|| key == 110) { return; }else continue; }ifnum = 0;if(key==224)key=getch( );switch (key){ case 75: left( ); break;case 77: right( ); break;case 72: up( ); break;case 80: down( );break;}if (score > best) { best = score; }if (ifnum) { adnum( ); show( ); }}}int jsk( )//计算空格数{ int n = 0;for (i = 0; i < 4; i++){ for (j = 0; j < 4; j++) { if ( a[i][j] == 0) {n++;} } }return n;}void left( )//向左移动{ for (i = 0; i < 4; i++){for (j = 1, k = 0; j < 4; j++){ if (a[i][j] > 0){ if ( a[i][k] == a[i][j]){ a[i][k] *= 2; k++;score = score + 2 * a[i][j];a[i][j] = 0; ifnum = 1; }else if ( a[i][k] == 0) { a[i][k] = a[i][j]; a[i][j] = 0; ifnum = 1; }else { a[i][k + 1] = a[i][j]; if ((k + 1) != j) { a[i][j] = 0; ifnum = 1; }k++; }}}}}void right( )//向右移动{for (i = 0; i < 4; i++){for (j = 2, k = 3; j >= 0; j--){if (a[i][j] > 0){ if (a[i][k] == a[i][j]){a[i][k] *= 2; k--; score = score + 2 * a[i][j]; a[i][j] = 0; ifnum = 1; }else if ( a[i][k] == 0) {a[i][k] = a[i][j]; a[i][j] = 0; ifnum = 1; }else { a[i][k - 1] = a[i][j]; if ((k - 1) != j) { a[i][j] = 0; ifnum = 1; } k--; }}}}}void up( )//向上移动{for (i = 0; i < 4; i++){for (j = 1, k = 0; j < 4; j++){if (a[j][i] > 0){if ( a[k][i] == a[j][i]) { a[k][i] *= 2; k++;score = score + 2 * a[j][i];a[j][i] = 0; ifnum = 1; }else if ( a[k][i] == 0) { a[k][i] = a[j][i]; a[j][i] = 0; ifnum = 1; }else { a[k + 1][i] = a[j][i]; if ((k + 1) != j) { a[j][i] = 0; ifnum = 1; }k++; }}}}}void down( )//向下移动{ for (i = 0; i < 4; i++){for (j = 2, k = 3; j >= 0; j--){if (a[j][i] > 0){if (a[k][i] == a[j][i]){a[k][i] *= 2; k--;score = score + 2 * a[j][i]; a[j][i] = 0; ifnum = 1; }else if (a[k][i] == 0) {a[k][i] = a[j][i]; a[j][i] = 0; ifnum = 1; }else {a[k - 1][i] = a[j][i];if ((k - 1) != j) {a[j][i] = 0; ifnum = 1; } k--; }}}}}void adnum( )//添加随机数{ srand(time(0)); //启动随机数发生器int n = rand( ) % jsk( );for (int i = 0; i < 4; i++){for (int j = 0; j < 4; j++){ if (a[i][j] == 0) { if (n != 0) { n--; }else {int k = rand() % 3;if (k == 0 || k == 1) {a[i][j] = 2; return; }else {a[i][j] = 4; return; } }}}}}void yes( )//游戏是否结束{ for (int i = 0; i < 4; i++){for (int j = 0; j < 3; j++){if (a[i][j] == a[i][j + 1] || a[j][i] == a[j + 1][i]) {over = 0; return; }}}over = 1;}void gtxy(int x, int y)//控制光标位置的函数{ COORD coord;coord.X = x;coord.Y = y;SetConsoleCursorPosition(GetStdHandle(STD_OUTPUT_HANDLE), coord);}
程序源码及运行结果如下所示(系统会自动去掉句首空格,排版较乱请见谅):#includeint main(){int max,min;int sum=0;printf("请输入一个整数:");scanf("%d",&max);printf("请再输入一个整数:");scanf("%d",&min);if(min>max){int temp;temp = max;max = min;min = temp;}for(int i=min;i<=max;i++){sum += i;}printf("两数之间的整数和是:%dn",sum);return 0;}
#include int fun(int num1, int num2);void main(){ int num1, num2; int sum; printf("请输入两个整数:n"); scanf("%d %d", &num1, &num2); if(num1 > num2) { int tmp = num1; num1 = num2; num2 = tmp; } sum = fun(num1, num2); printf("大于等于%d小于等于%d的所有整数和是%dn", num1, num2, sum);}int fun(int num1, int num2){ int i, sum = 0; for(i=num1; i<=num2; i++) { sum += i; } return sum;}

c语言必背代码有哪些?
1、/*输出9*9口诀。共9行9列,i控制行,j控制列。*/#include "stdio.h"main(){int i,j,result;for (i=1;i<10;i++){ for(j=1;j<10;j++){result=i*j;printf("%d*%d=%-3d",i,j,result);/*-3d表示左对齐,占3位*/}printf("n");/*每一行后换行*/}}2、/*古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?兔子的规律为数列1,1,2,3,5,8,13,21....*/main(){long f1,f2;int i;f1=f2=1;for(i=1;i<=20;i++){ printf("%12ld %12ld",f1,f2);if(i%2==0) printf("n");/*控制输出,每行四个*/f1=f1+f2; /*前两个月加起来赋值给第三个月*/f2=f1+f2; /*前两个月加起来赋值给第三个月*/}}3、/*判断101-200之间有多少个素数,并输出所有素数及素数的个数。程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,则表明此数不是素数,反之是素数。*/#include "math.h"main(){int m,i,k,h=0,leap=1;printf("n");for(m=101;m<=200;m++){ k=sqrt(m+1);for(i=2;i<=k;i++)if(m%i==0){leap=0;break;}if(leap) /*内循环结束后,leap依然为1,则m是素数*/{printf("%-4d",m);h++;if(h%10==0)printf("n");}leap=1;}printf("nThe total is %d",h);}4、/*一个数如果恰好等于它的因子之和,这个数就称为"完数"。例如6=1+2+3.编程找出1000以内的所有完数。*/main(){static int k[10];int i,j,n,s;for(j=2;j<1000;j++){n=-1;s=j;for(i=1;i#include main(){ float a[4][5],sum1,sum2;int i,j;for(i=0;i<3;i++)for(j=0;j<4;j++)scanf("%f",&a[i][j]);for(i=0;i<3;i++){ sum1=0;for(j=0;j<4;j++)sum1+=a[i][j];a[i][4]=sum1/4;}
c语言必背的九个经典代码如下:1、/*输出9*9口诀。共9行9列,i控制行,j控制列。*/#include"stdio.h"main(){inti,j,result;for (i=1;i<10;i++){ for(j=1;j<10;j++){result=i*j;printf("%d*%d=%-3d",i,j,result);/*-3d表示左对齐,占3位*/}printf("n");/*每一行后换行*/}system("pause");}2、/*古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?兔子的规律为数列1,1,2,3,5,8,13,21....*/main(){long f1,f2;int i;f1=f2=1;for(i=1;i<=20;i++){ printf("%12ld %12ld",f1,f2);if(i%2==0) printf("n");/*控制输出,每行四个*/f1=f1+f2; /*前两个月加起来赋值给第三个月*/f2=f1+f2; /*前两个月加起来赋值给第三个月*/}}3、/*判断101-200之间有多少个素数,并输出所有素数及素数的个数。程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,则表明此数不是素数,反之是素数。*/#include"math.h"main(){int m,i,k,h=0,leap=1;printf("n");for(m=101;m<=200;m++){ k=sqrt(m);for(i=2;i<=k;i++)if(m%i==0){leap=0;break;}if(leap) /*内循环结束后,leap依然为1,则m是素数*/{printf("%-4d",m);h++;if(h%10==0)printf("n");}leap=1;}printf("nThe total is %d",h);system("pause");}4、/*一个数如果恰好等于它的因子之和,这个数就称为“完数”。例如6=1+2+3.编程找出1000以内的所有完数。*/main(){static int k[10];int i,j,n,s;for(j=2;j<1000;j++){n=-1;s=j;for(i=1;i#includemain(){ floata[4][5],sum1,sum2;int i,j;for(i=0;i<3;i++)for(j=0;j<4;j++)scanf("%f",&a[i][j]);for(i=0;i<3;i++){ sum1=0;for(j=0;j<4;j++)sum1+=a[i][j];a[i][4]=sum1/4;}for(j=0;j<5;j++){ sum2=0;for(i=0;i<3;i++)sum2+=a[i][j];a[3][j]=sum2/3;}for(i=0;i<4;i++){ for(j=0;j<5;j++)printf("%6.2f",a[i][j]);printf("n");}system("pause");}8、/*完善程序,实现将输入的字符串反序输出,如输入windows 输出swodniw。*/#includemain(){ char c[200],c1;int i,j,k;printf("Enter a string: ");scanf("%s",c);k=strlen(c);for (i=0,j=k-1;imain(){ char s[80],c;int j,k;printf("nEnter a string: ");gets(s);printf("nEnter a character: ");c=getchar( );for(j=k=0;s[j]!= ' ';j++)if(s[j]!=c)s[k++]=s[j];s[k]= ' ';printf("n%s",s);system("pause");}
c语言必背代码:1、/*输出9*9口诀。共9行9列,i控制行,j控制列。*/#include "stdio.h"main(){int i,j,result;for (i=1;i<10;i++){ for(j=1;j<10;j++){result=i*j;printf("%d*%d=%-3d",i,j,result);/*-3d表示左对齐,占3位*/}printf(" ");/*每一行后换行*/}}2、/*完善程序,实现将输入的字符串反序输出,如输入windows 输出swodniw。*/#include main(){ char c[200],c1;int i,j,k;printf("Enter a string: ");scanf("%s",c);k=strlen(c);for (i=0,j=k-1;ivoid sort(int *x,int n){int i,j,k,t;for(i=0;ix[k]) k=j;if(k!=i){t=x[i];x[i]=x[k];x[k]=t;}}}void main(){FILE *fp;int *p,i,a[10];fp=fopen("p9_1.out","w");p=a;printf("Input 10 numbers:");for(i=0;i<10;i++)scanf("%d",p++);p=a;sort(p,10);for(;preplace(char *s,char c1,char c2){ while(*s!=''){ if (*s==c1)*s=c2;s++;}}main(){ FILE *fp;char str[100],a,b;if((fp=fopen("p10_2.out","w"))==NULL){ printf("cannot open the file ");exit(0); }printf("Enter a string: ");gets(str);printf("Enter a&&b: ");scanf("%c,%c",&a,&b);printf("%s ",str);fprintf(fp,"%s ",str);replace(str,a,b);printf("The new string is----%s ",str);fprintf(fp,"The new string is----%s ",str);fclose(fp);}5、/*在一个字串s1中查找一子串s2,若存在则返回子串在主串中的起始位置,不存在则返回-1。*/main(){char s1[6]="thisis";char s2[5]="is";printf("%d ",search(s1,s2));system("pause");}int search(char s1[],char s2[]){int i=0,j,len=strlen(s2);while(s1[i]){for(j=0;j=len)return i;else i++;}return -1;}6、/*建立一个有三个结点的简单链表:*/#define NULL 0struct student{int num;char *name;int age ;struct student *next;};void main(){struct student a,b,c,*head,*p;a.num=1001; a.name="lihua"; a.age=18; /* 对结点成员进行赋值 */b.num=1002; b.name="liuxing"; b.age=19;c.num=1003; c.name="huangke"; c.age=18;head=&a; /* 建立链表,a为头结点 */a.next=&b;b.next=&c;c.next=NULL;p=head; /* 输出链表 */do{printf("%5d,%s,%3d ",p->num,p->name,p->age);p=p->next;}while(p!=NULL);}
C语言必背18个经典程序1、#include "stdio.h"main(){int i,j,result;for (i=1;i<10;i++){ for(j=1;j<10;j++){result=i*j;printf("%d*%d=%-3d",i,j,result);}printf("n");}}2、main(){long f1,f2;int i;f1=f2=1;for(i=1;i<=20;i++){ printf("ld ld",f1,f2);if(i%2==0) printf("n");f1=f1+f2;f2=f1+f2;}}3、#include "math.h"main(){int m,i,k,h=0,leap=1;printf("n");for(m=101;m<=200;m++){ k=sqrt(m+1);for(i=2;i<=k;i++)if(m%i==0){leap=0;break;}if(leap){printf("%-4d",m);h++;if(h==0)printf("n");}leap=1;}printf("nThe total is %d",h);}4、main(){static int k[10];int i,j,n,s;for(j=2;j<1000;j++){n=-1;s=j;for(i=1;i{if((j%i)==0){ n++;s=s-i;k[n]=i;}}if(s==0){printf("%d is a wanshu: ",j);for(i=0;iprintf("%d,",k[i]);printf("%dn",k[n]);}}}5、main(){ int a[4][4],b[4][4],i,j;printf("input 16 numbers: ");for(i=0;i<4;i++)for(j=0;j<4;j++){ scanf("%d",&a[i][j]);b[3-j][i]=a[i][j];}printf("array b:n");for(i=0;i<4;i++){ for(j=0;j<4;j++)printf("m",b[i][j]);printf("n");}}6、main(){int i,j,a[6][6];for(i=0;i<=5;i++){a[i][i]=1;a[i][0]=1;}for(i=2;i<=5;i++)for(j=1;j<=i-1;j++)a[i][j]=a[i-1][j]+a[i-1][j-1];for(i=0;i<=5;i++){for(j=0;j<=i;j++)printf("M",a[i][j]);printf("n");}}7、#include#includemain(){ float a[4][5],sum1,sum2;int i,j;for(i=0;i<3;i++)for(j=0;j<4;j++)scanf("%f",&a[i][j]);for(i=0;i<3;i++){ sum1=0;for(j=0;j<4;j++)sum1+=a[i][j];a[i][4]=sum1/4;}for(j=0;j<5;j++){ sum2=0;for(i=0;i<3;i++)sum2+=a[i][j];a[3][j]=sum2/3;}for(i=0;i<4;i++){ for(j=0;j<5;j++)printf("%6.2f",a[i][j]);printf("n");}}8、#includemain(){ char c[200],c1;int i,j,k;printf("Enter a string: ");scanf("%s",c);k=strlen(c);for (i=0,j=k-1;i{ c1=c[i];c[i]=c[j];c[j]=c1; }printf("%sn",c);}指针法:void invert(char *s){int i,j,k;char t;k=strlen(s);for(i=0,j=k-1;i{ t=*(s+i); *(s+i)=*(s+j); *(s+j)=t; }}main(){ FILE *fp;char str[200],*p,i,j;if((fp=fopen("p9_2.out","w"))==NULL){ printf("cannot open the filen");exit(0);}printf("input str:n");gets(str);printf(“n%s”,str);fprintf(fp,“%s”,str);invert(str);printf(“n%s”,str);fprintf(fp,“n%s”,str);fclose(fp);}9、#includemain(){ char s[80],c;int j,k;printf("nEnter a string: ");gets(s);printf("nEnter a character: ");c=getchar( );for(j=k=0;s[j]!= ' ';j++)if(s[j]!=c)s[k++]=s[j];s[k]= ' ';printf("n%s",s);}10、#includevoid sort(int *x,int n){int i,j,k,t;for(i=0;i{k=i;for(j=i+1;jif(x[j]>x[k]) k=j;if(k!=i){t=x[i];x[i]=x[k];x[k]=t;}}}void main(){FILE *fp;int *p,i,a[10];fp=fopen("p9_1.out","w");p=a;printf("Input 10 numbers:");for(i=0;i<10;i++)scanf("%d",p++);p=a;sort(p,10);for(;p{ printf("%d ",*p);fprintf(fp,"%d ",*p); }system("pause");fclose(fp);}11、已知数组a中的元素已按由小到大顺序排列,以下程序的功能是将输入的一个数插入数组a中,插入后,数组a中的元素仍然由小到大顺序排列*/main(){ int a[10]={0,12,17,20,25,28,30};int x , i, j=6;printf("Enter a number: ");scanf("%d",&x);a[0]=x;i=j;while(a[i]>x){ a[i+1]=a[i]; i--; }a[++i]=x;j++;for(i=1;i<=j;i++) printf("�",a[i]);printf("n");}12、#includereplace(char *s,char c1,char c2){ while(*s!=' '){ if (*s==c1)*s=c2;s++;}}main(){ FILE *fp;char str[100],a,b;if((fp=fopen("p10_2.out","w"))==NULL){ printf("cannot open the filen");exit(0); }printf("Enter a string:n");gets(str);printf("Enter a&&b:n");scanf("%c,%c",&a,&b);printf("%sn",str);fprintf(fp,"%sn",str);replace(str,a,b);printf("The new string is----%sn",str);fprintf(fp,"The new string is----%sn",str);fclose(fp);}13、main(){char s1[6]="thisis";char s2[5]="is";printf("%dn",search(s1,s2));system("pause");}int search(char s1[],char s2[]){int i=0,j,len=strlen(s2);while(s1[i]){for(j=0;jif(s1[i+j]!=s2[j]) break;if(j>=len)return i;else i++;}return -1;}14、struct student{int num;char *name;char sex;int age;}stu[5]={{1001,"lihua",'F',18},{1002,"liuxing",'M',19},{1003,"huangke",'F',19},{1004,"fengshou",'F',19},{1005,"Wangming",'M',18}};main(){int i;struct student *ps;printf("Num tNametttSextAgetn");for(ps=stu;psprintf("%dt%-10stt%ct%dtn",ps->num,ps->name,ps->sex,ps->age);for(i=0;i<5;i++)printf("%dt%dtn",stu[i].num,stu[i].age);}15、#define NULL 0struct student{int num;char *name;int age ;struct student *next;};void main(){struct student a,b,c,*head,*p;a.num=1001; a.name="lihua"; a.age=18;b.num=1002; b.name="liuxing"; b.age=19;c.num=1003; c.name="huangke"; c.age=18;head=&a;a.next=&b;b.next=&c;c.next=NULL;p=head;do{printf("],%s,=n",p->num,p->name,p->age);p=p->next;}while(p!=NULL);}16、#include#include#includemain(){ char s[100];int i,j,n;printf("输入字符串:n");gets(s);n=strlen(s);for(i=0,j=n-1;iif(s[i]!=s[j]) break;if(i>=j) printf("是回文串n");else printf("不是回文串n");}17、#includevoid fun(int a[],int n){int i,j,t;for(i=0;i<=n-1;i++)for(j=0;jif(a[j]>a[j+1]) {t=a[j];a[j]=a[j+1];a[j+1]=t;}}main(){int a[10]={12,45,7,8,96,4,10,48,2,46},n=10,i;FILE *f;if((f=fopen("myf2.out","w"))==NULL)printf("open file myf2.out failed!n");fun(a,10);for(i=0;i<10;i++){printf("M",a[i]);fprintf(f,"M",a[i]);}fclose(f);}18、编写函数countpi,利用公式计算π的近似值,当某一项的值小于10-5时,认为达到精度要求,请完善函数。将结果显示在屏幕上并输出到文件p7_3.out中。#includedouble countpi(double eps){int m=1;double temp=1.0,s=0;while(temp>=eps){ s+=temp;temp=temp*m/(2*m+1);m++;}return(2*s);}main(){FILE *fp;double eps=1e-5,pi;if((fp=fopen("p7_3.out","w"))==NULL){ printf("cannot open the filen");exit(0);}pi= countpi(eps);printf("pi=%lfn",pi);fprintf(fp,"pi=%lfn",pi);fclose(fp);}
1、/*古典问题:有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问每个月的兔子总数为多少?兔子的规律为数列1,1,2,3,5,8,13,21....*/main(){long f1,f2;int i;f1=f2=1;for(i=1;i<=20;i++){ printf("%12ld%12ld",f1,f2);if(i%2==0) printf("n");/*控制输出,每行四个*/f1=f1+f2;/*前两个月加起来赋值给第三个月*/f2=f1+f2; /*前两个月加起来赋值给第三个月*/}}2、/*判断101-200之间有多少个素数,并输出所有素数及素数的个数。程序分析:判断素数的方法:用一个数分别去除2到sqrt(这个数),如果能被整除,则表明此数不是素数,反之是素数。*/#include "math.h"main(){intm,i,k,h=0,leap=1;printf("n");for(m=101;m<=200;m++){k=sqrt(m+1);for(i=2;i<=k;i++)if(m%i==0){leap=0;break;}if(leap) /*内循环结束后,leap依然为1,则m是素数*/{printf("%-4d",m);h++;if(h%10==0)printf("n");}leap=1;}printf("nThetotal is %d",h);}3、/*一个数如果恰好等于它的因子之和,这个数就称为“完数”。例如6=1+2+3.编程找出1000以内的所有完数。*/main(){static int k[10];inti,j,n,s;for(j=2;j<1000;j++){n=-1;s=j;for(i=1;i#include main(){ float a[4][5],sum1,sum2;inti,j;for(i=0;i<3;i++)for(j=0;j<4;j++)scanf("%f",&a[i][j]);for(i=0;i<3;i++){sum1=0;for(j=0;j<4;j++)sum1+=a[i][j];a[i][4]=sum1/4;}for(j=0;j<5;j++){ sum2=0;for(i=0;i<3;i++)sum2+=a[i][j];a[3][j]=sum2/3;}for(i=0;i<4;i++){ for(j=0;j<5;j++)printf("%6.2f",a[i][j]);printf("n");}}
