题目地址
自测-1
#include<stdio.h>
int main(int argc,char** argv)
{
int sum;
char ch;
scanf("%d %c",&sum,&ch);
int i=1,msum=1,fk=sum-msum;
while(fk>=(i+2)*2)
{
i+=2;
msum+=2*i;
fk=sum-msum;
}
int ins=i;
while(ins>1)
{
int ind=ins,inf=(i-ins)/2;
while(inf--)
putchar(' ');
while(ind--)
putchar(ch);
putchar('\n');
ins-=2;
}
int inf=(i-1)/2;
while(inf--)
putchar(' ');
putchar(ch);
putchar('\n');
ins=3;
while(ins<=i)
{
int ind=ins,inf=(i-ins)/2;
while(inf--)
putchar(' ');
while(ind--)
putchar(ch);
putchar('\n');
ins+=2;
}
printf("%d\n",fk);
return 0;
}>>先计算出最大可以容下多少个符号和最大值时第一行符号个数，后循环输出空格和符号即可。
自测-2
#include<stdio.h>
#include<math.h>
int book[100005];
int isprime(int m)
{
int i;
if(m<2)
return 0;
for(i=2;i<=sqrt(m);i++)
if(!(m%i))
return 0;
return 1;
}
int main(int argc,char** argv)
{
int N;
scanf("%d",&N);
int i,sum=0;
for(i=2;i<=N;i++)
book[i]=isprime(i);
for(i=4;i<=N;i++)
if(book[i]&&book[i]-book[i-2]==0)
sum++;
printf("%d\n",sum);
return 0;
}>>标记2~N所有的素数，检查计数相差为2的素数。
自测-3
#include<stdio.h>
int main(int argc,char** argv)
{
int m,n;
scanf("%d%d",&m,&n);
int i,dig[105];
for(i=0;i<m;i++)
scanf("%d",&dig[i]);
n=m-n;
while(n<0)
n+=m;
for(i=1;i<=m;i++)
{
n%=m;
printf("%d%c",dig[n++],i==m?'\n':' ');
}
return 0;
}>>每次递增都取余运算，计算初始位置 m-n 时可能为负数，须加 m 直到是正数为止。
自测-4
#include<stdio.h>
#include<string.h>
char str[25];
int dig[10];
int main(int argc,char** argv)
{
scanf("%s",str);
int i,check=0;
for(i=0;i<strlen(str);i++)
{
if(!dig[str[i]-'0'])
check++;
dig[str[i]-'0']++;
}
int num=0;
for(i--;i>=0;i--)
{
int fk=num;
num=(str[i]-'0')*2;
str[i]=(num+fk)%10+'0';
num/=10;
}
if(num)
printf("No\n%d",num);
else
{
for(i=0;i<strlen(str);i++)
{
dig[str[i]-'0']--;
if(!dig[str[i]-'0'])
check--;
}
if(check)
printf("No\n");
else
printf("Yes\n");
}
printf("%s\n",str);
return 0;
}>>1.数组存储后先逐位记录用到的数字和个数；
2.每一位乘二后进位；
3.如果计算后存在进位不为 0 ，直接输出 No 并输出当前进位；
4.否则检查当前数组是否和最初用到的数字相同、个数一样，一样输出 Yes ，不一样输出 No；
5.输出当前数组。
自测-5
#include<stdio.h>
#include<string.h>
char str[55][5]={"x","S1","S2","S3","S4","S5","S6","S7","S8","S9","S10","S11","S12","S13",\
"H1","H2","H3","H4","H5","H6","H7","H8","H9","H10","H11","H12","H13",\
"C1","C2","C3","C4","C5","C6","C7","C8","C9","C10","C11","C12","C13",\
"D1","D2","D3","D4","D5","D6","D7","D8","D9","D10","D11","D12","D13",\
"J1","J2"};
int dig[55];
int fkkk[55];
int main(int argc,char** argv)
{
int k;
scanf("%d",&k);
int i,j;
for(i=1;i<=54;i++)
scanf("%d",&dig[i]);
for(i=1;i<=54;i++)
{
int sp=dig[i];
for(j=2;j<=k;j++)
sp=dig[sp];
fkkk[sp]=i;
}
for(i=1;i<=54;i++)
printf("%s%c",str[fkkk[i]],i==54?'\n':' ');
return 0;
}>>计算洗牌 k 次后的各牌顺序，fkkk[ i ] 数组存储洗牌完成后的第 i 张牌yu原来是第第几张牌。
这个过程有点像用数组实现链表。
扫描二维码关注公众号，回复：
2598004 查看本文章
