每日三百行代码 第二十二天
#include<stdio.h>
int main()
{
int i,j=1,x=0,y=0,rest,N;
char C;
scanf("%d %c",&N,&C);
while(2*j*j-1<=N)//计算层数
{
j++;
}
j--;
y=2*j-1;
rest=N-2*j*j+1;//剩下字符的数量
while(y>0)//输出沙漏上半部分
{
for(i=0;i<x;i++)
printf(" ");
for(i=0;i<y;i++)
printf("%c",C);
printf("\n");
x++;
y-=2;
}
x--;
y+=2;
while(x>0)//输出沙漏下半部分
{
x--;
y+=2;
for(i=0;i<x;i++)
printf(" ");
for(i=0;i<y;i++)
printf("%c",C);
printf("\n");
}
printf("%d",rest);
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,i,j,flag,k,sum=0,w=0;
int a[112345];
a[0]=2;
scanf("%d",&n);
for(i=3; i<=n; i++)
{
flag=1;
k=sqrt(i);
for(j=2; j<=k; j++){
if(i%j==0){
flag=0;
break;
}
}
if(flag){
a[++w]=i;
}
}
for(i=0; i<=w; i++)
{
if(a[i+1]-a[i]==2)
{
sum++;
}
}
printf("%d",sum);
}
#include<bits/stdc++.h>
using namespace std;
bool judge(int n)
{
if(n<=1) return false;
int sqrtN=sqrt(1.0*n);
for(int i=2;i<=sqrtN;i++)
{
if(n%i==0) return false;
}
return true;
}
int main()
{
int n,cnt=0;
scanf("%d",&n);
for(int i=2;i<=n;i++)
{
if(judge(i)&&judge(i-2))
cnt++;
}
printf("%d\n",cnt);
return 0;
}
#include <stdio.h>
int main() {
int n, m;
int i, j;
int k;
scanf("%d %d", &n, &m);
int a[n];
for (i = 0; i < n; i++) {
scanf("%d", &a[i]);
}
for (j = 0; j < m; j++) {
k = a[n - 1];
for (i = n - 1; i > 0; i--) {
a[i] = a[(i - 1 + n) % n];
}
a[0] = k;
}
for (i = 0; i < n; i++) {
if (i == n - 1)
printf("%d", a[i]);
else
printf("%d ", a[i]);
}
printf("\n");
return 0;
}
翻译：请注意，编号 123456789 是一个 9 位数字，正好由 1 到 9 的数字组成，没有重复。翻倍，我们将获得246913578，这恰好是另一个9位数字，正好由1到9的数字组成，只有在不同的排列中。检查，看看结果，如果我们再次加倍！
现在，你应该检查是否有更多的数字与这个属性。即将给定数字翻倍K数字，你要告诉，如果由此产生的数字只包括数字在原始数字的排列。
输入规范：
每个输入包含一个测试案例。每个案例包含一个不超过 20 位数字的正整数。
输出规格：
对于每个测试案例，如果输入数字翻倍仅包含原始数字中数字的排列，则首先打印在行"是"中，如果没有，则第一次打印为"否"。然后在下一行中，打印双倍的数字。
#include <stdio.h>
#include <string.h>
//大数乘法
#define MAX 30
char num[MAX];
int res[MAX]={
0};
//这里易错，要11位
char a1[11] = "0000000000";
char a2[11] = "0000000000";
int main() {
//方便输入
scanf("%s", num);
int len = strlen(num);
int temp, carry = 0, index;
for(int j=len-1, m=0; len>=0; --len, --j, ++m) {
//这个分支保证可以最高位可以进位
if(len==0) {
temp = carry;
}
else {
temp = (int)(num[j]-'0')*2 + carry;
}
res[m] = temp%10;
carry = temp/10;
}
//去除前导0，得到index
for(index=MAX-1; index>=0; --index) {
if(res[index]!=0) break;
}
//下面两个循环比较，桶~~
for(int i=0; i<strlen(num); ++i) {
++a1[(int)(num[i]-'0')];
}
for(int i=0; i<=index; ++i) {
++a2[res[i]];
}
//注意考虑0的情况
if(strcmp(a1,a2)==0 || strcmp(num, "0")==0)
printf("Yes\n");
else
printf("No\n");
if(strcmp(num, "0")==0) {
printf("0");
}
else {
while(index>=0) {
printf("%d", res[index]);
--index;
}
}
return 0;
}
#include<stdio.h>
int main()
{
int ch1[21], ch2[22] = {
0 };
char temp;
int i = 0, num[10] = {
0 };
while ((temp = getchar()) != '\n')
{
ch1[i] = temp - '0';
i++;
num[(temp - '0')]++;
}
int i2 = 21;
ch2[i2] = (ch1[i-1] * 2) % 10;
i2--;
int j = i - 2;
for (; j >= 0; j--)
{
int buf = ch1[j + 1] * 2 / 10 + ch1[j] * 2 % 10;
ch2[i2] = buf;
i2--;
}
ch2[i2] = ch1[j + 1] * 2 / 10;
if(ch2[i2]!=0)
i2--;
for (int k = 21; k > i2; k--)
{
num[ch2[k]]--;
}
int flag = 0;
for (int i = 0; i < 10; i++)
{
if (num[i] != 0)
{
printf("No\n");
flag = 1;
break;
}
}
if(flag==0)
printf("Yes\n");
int k = 0;
while (ch2[k] == 0)
{
k++;
}
for (; k < 22; k++)
{
printf("%d", ch2[k]);
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int ht[10]={
0};
string DoubleS(string str)
{
int temp=0,r,carry=0;
string result="";
for(int i=str.size()-1;i>=0;i--)
{
ht[str[i]-'0']++;
r=(str[i]-'0')*2+carry;
carry=r/10;
ht[r%10]--;
result+=(r%10+'0');
}
if(carry!=0)
{
result+=(carry+'0');
ht[carry]--;
}
reverse(result.begin(),result.end());
return result;
}
int main()
{
string ori,res;
cin>>ori;
res=DoubleS(ori);
bool flag=true;
for(int i=0;i<res.size()&&flag;i++)
{
if(ht[i]!=0) flag=false;
}
if(flag) cout<<"Yes\n";
else cout<<"No\n";
cout<<res<<endl;
return 0;
}
#include<stdio.h>
#include<string.h>
int main()
{
char card1[54][4] = {
"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" }, card2[54][4] = {
};
int n, shuffle[54];
scanf("%d\n", &n);
for (int i = 0; i < 54; i++)
{
scanf("%d", &shuffle[i]);
}
for (int i = 1; i <= n; i++)
{
if (i % 2 == 1)
{
for (int j = 0; j < 54; j++)
{
strcpy(card2[shuffle[j] - 1], card1[j]);
}
}
if (i % 2 == 0)
{
for (int j = 0; j < 54; j++)
{
strcpy(card1[shuffle[j] - 1], card2[j]);
}
}
}
if (n % 2 == 1)
{
printf("%s", card2[0]);
for (int i = 1; i < 54; i++)
printf(" %s", card2[i]);
}
else
{
printf("%s", card1[0]);
for (int i = 1; i < 54; i++)
printf(" %s", card1[i]);
}
return 0;
}
#include <stdlib.h>
#include <stdio.h>
int main(){
int a[55],b[55],c[55];
int i,j,count,m,n;
scanf("%d",&count);
for(i=0;i<54;i++){
a[i] = i+1;
//printf("%d ",a[i]); //创建1-54的数字列表a，对应54张卡牌
}
//printf("\n");
for(i=0;i<54;i++){
scanf("%d",&b[i]); //输入排列顺序，存储在b数组内
//printf("%d ",b[i]);
}
//printf("\n");
for(i=0;i<54;i++){
b[i] = b[i]-1; //顺序数全部减一，方便对应数组中的下标值
//printf("%d ",b[i]);
}
//printf("\n");
for(j=0;j<count;j++){
for(i=0;i<54;i++){
c[b[i]]=a[i]; //循环赋值count次，得出洗牌后的卡牌数值
}
for(i=0;i<54;i++){
a[i] = c[i];
}
}
for(i=0;i<54;i++){
//printf("%d ",c[i]);
m = c[i]/13;
n = c[i]%13;
if(n==0){
n=13;
m--;
}
if(m==0){
printf("S%d",n);
}else if(m==1){
printf("H%d",n);
}else if(m==2){
printf("C%d",n);
}else if(m==3){
printf("D%d",n);
}else if(m==4){
printf("J%d",n);
}
if(i != 53){
printf(" ");
}
}
return 0;
}
//zc-5
#include <stdio.h>
typedef struct PK{
char c;
int i;
}P;
int main ( void )
{
int k, i, j;
P p[54];
scanf( "%d", &k );
int a[54];
for( i = 0; i < 54; i++ )
scanf( "%d", &a[i] );
for( i = 0; i < 54; i++ ){
if( i < 13 ){
p[i].c = 'S';
p[i].i = i % 13 + 1;
}
else if( i < 26 ){
p[i].c = 'H';
p[i].i = i % 13 + 1;
}
else if( i < 39 ){
p[i].c = 'C';
p[i].i = i % 13 + 1;
}
else if( i < 52 ){
p[i].c = 'D';
p[i].i = i % 13 + 1;
}
else{
p[i].c = 'J';
p[i].i = i % 13 + 1;
}
}
for( j = 0; j < k; j++ )
{
P s1[54];
for( i = 0; i < 54 ; i++ )
s1[ a[i] - 1 ] = p[i];
for( i = 0; i < 54; i++ )
p[i] = s1[i];
}
for( i = 0; i < 54; i++ ){
if( i != 53 )
printf( "%c%d ", p[i].c, p[i].i );
else
printf( "%c%d", p[i].c, p[i].i );
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
char s1[5]={
'S','H','C','D','J'};
int main(){
int card1[55],n,operate[55],card2[55]={
0};
for(int i=1;i<55;i++){
card1[i]=i;
}
scanf("%d",&n);
for(int i=1;i<55;i++){
scanf("%d",&operate[i]);
}
for(int i=1;i<=n;i++){
for(int j=1;j<55;j++){
card2[operate[j]]=card1[j];
}
for(int k=1;k<55;k++){
card1[k]=card2[k];
}
}
for(int i=1;i<55;i++){
if(i!=1){
printf(" ");
}
card2[i]--;
printf("%c%d",s1[card2[i]/13],card2[i]%13+1);
}
return 0;
}
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