1.编写程序，将一个整数逆置后输出如1234逆置后为4321。main()int n,a,s=0;scanf(%d,&n);while (n)a=n%10;s=s*10+a;n=n/10;printf(%d,s);2.统计字符串当中数字字符出现的次数#include stdio.hmain()int n10=0,i=0;char s81;gets(s);while(si) if(si=0 &si=9)nsi-0+;i+;for(i=0;i10;i+)printf(%3d,ni);3.编写程序，计算n*n阶矩阵周边元素绝对值之和。main()int a44,i,j,sum;sum=0;for(i=0;i4;i+)for(j=0;j4;j+)scanf(%d,*(a+i)+j);if(aij0)aij=-aij;for(i=0;i4;i+)for(j=0;j4;j+)if (i=0|i=3|j=0|j=3)sum +=* (*(a+i)+j);printf(sum=%dn,sum);4输出所有的4位回文数的个数。main() int i,j,k,w,sum;sum=0;for(i=1;i=9;i+)for(j=0;j=9;j+)for(k=0;k=9;k+)for(w=0;w=0;i-) printf(%d,ai);6.编写程序，统计一个英文句子中最长的单词的字符数。# include stdio.hfun (char *s)int n=0,m; while(*s) while(*s= )s+; m=0; while(*s&*s!= ) m+;s+; if(mn)n=m; return n; main()char s81; int n; gets(s); n=fun(s); printf(%d,n);7编写程序，将一个无字符整形数转换成其对应的数字串#include stdio.hmain() int a20,i=0,k; int s,t,j; scanf(%d,&s); while(s) ai+=s%10; s/=10; k=i-1; for(i=0,j=k;ij;i+,j-) t=ai;ai=aj;aj=t; for(i=0;i=k;i+) print(%d,ai);8编写程序，将整型数组a中的前k个元素逆置#include stdio.h#define n 4main()int a10,i,j,t;for(i=0;i10;i+)scanf(%d,&ai);for(i=0,j=n;ij;i+,j+)t=ai;ai=aj;aj=t;for(i=0;i10;i+)printf(%4d,ai);8.有问题9编写程序，统计子串t在主串s中出现的次数fun(char *s,char *t)char *p1,*p2;int n=0; while(*s&*t) p1=s,p2=t; while(*p1&*p2) if(*p1=*p2) p1+;p2+; else break; if(*p2=0) n+;s=p1; else s+; return n;#include stdio.h main() char s81,t20; int n; gets(s); gets(t); n=fun(s,t); printf(%d,n); 10.统计子串中t在主串s中出现的次数char *fun(char *s,char *t) char *p1,*p2; int *k; while(*s&*t) p1=s; p2=t; while(*p1&*p2) if(*p1=*p2) p1+; p2+; else break; if(*p2=0) k=s;s=p1; else s+; return k;# include stdio.hmain()char s81,t20; char *n; gets(s); gets(t); n=fun(s,t); printf(%d,n-s);11.计算N*n接矩阵主对角线元素生序排序#include stdio.h#define n 4main()int ann,i,j,sum=0;printf(input array a(n*n):n);for(i=0;in;i+)for(j=0;jn;j+)scanf(%d,&aij);if(aij0) aij=-aij;for(i=0;in;i+) for(j=0;jn;j+) if(i=j|i+j=n-1) sum+=aij; printf(sum=%dn,sum); 23统计字符串中含有空格字符的个数#include stdio.hmain() int sum=0, i=0; char s81; gets(s); while(si) if (si= ) sum+=1; i+; for(i=0;ia&*p=A&*p=A&*p=a&*p=a & *p=z) *p-=32; word=1; p+; puts(a);39编写程序，求一为数组的最大值及其位置#define n 10main()int an ,i,max,pos; for(i=0;in;i+) scanf(%d, &ai); max=a0,pos=0; for(i=0;in;i+) if(max=A&si=z)nsi-A+;i+;for(i=0;i=a&si=z)nsi-a+;i+;for(i=0;i26;i+)printf(%3d,ni);30统计10009999之间满足个位数+十位数+百位数+千位数=10#include stdio.hmain() int a,b,c,d;int sum=0;for(a=1;a=9;a+)for(b=1;b=9;b+)for(c=1;c=9;c+)for(d=1;d=9;d+)if(a+b+c+d=10)sum+=1;printf(%d,sum);31统计10009999之间满足个位数+十位数=百位数+千位数#include stdio.hmain() int a,b,c,d;int sum=0;for(a=1;a=9;a+)for(b=1;b=9;b+)for(c=1;c=9;c+)for(d=1;d=9;d+)if(a+b=c+d