Openjudge 1.7 String

1.7.9 After the letter shift

Read one line of the code:

while((k=getchar())!=EOF)

---

ASCII code

Lowercase \ (A \) 97

Lowercase \ (Z \) 122

Uppercase \ (A \) 65

Uppercase \ (the Z-\) 90

Digital 0 \ (48 \)

Number 9 \ (57 \)

//1.7.9
#include<bits/stdc++.h>
using namespace std; 
int main()
{
    char p[85];
    char k;
    int i=0;
    while((k=getchar())!=EOF){
        if(k>=65&&k<=89) k++;
        else if(k==90) k=65;
        else if(k>=97&&k<=121) k++;
        else if(k==122) k=97;
        //else if(k==32) putchar('32');
        p[i]=k;
        i++;
        putchar(k); 
    }
    //puts(p);
    return 0;
}

1.7.11 Lurker

//1.7.11?
#include <algorithm>
#include <iostream>
#include <cmath>
#include <cstdio>
#include <cstring>
using namespace std;
#define N 1020
char s1[N],s2[N],s3[N];
int a[26];
bool fail = 0;
int main(){
    scanf("%s%s%s",s1,s2,s3);
    for(int i=0;i<26;i++){
        a[i]=-1;
    }
    int len=strlen(s1);
    if(len<26) fail=1;
    else{
        for(int i=0;i<len;i++){
            if(a[s1[i]-'A'] == -1){
                a[s1[i]-'A']=s2[i]- 'A';
            }else{
                if(a[s1[i]-'A']!= s2[i]- 'A'){
                    fail = 1;
                    break;
                }
            }
        }
        for(int i = 0;i<26;i++){
            if(a[i]==-1){
                fail = 1;
                break;
            }
        }
        for(int i = 0;i<26;i++){
            for(int j=i+1;j<26;j++){
                if(a[i]==a[j]){
                    fail = 1;
                    break;
                }
            }
        }
    }
    if(fail == 1){
      cout<<"Failed"<<endl;
    }
    else{
        len = strlen(s3);
        for(int i = 0; i <len;++i)
            s3[i]=a[s3[i]-'A']+'A';
        printf("%s\n",s3);
    }
    return 0;
}

1.7.18 validation substring

#include <cstirng>
  strstr(a,b);

The return value is 0 or 1, for determining whether a substring

//1.7.18
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
char a[205],b[205];

int main( ){
    gets(a);
    gets(b);
    int len1,len2;
    len1=strlen(a);len2=strlen(b);
    bool op=0;
    if(len2>len1){
        swap(a,b);
        swap(len1,len2);
    }
    if(strstr(a,b)) printf("%s is substring of %s",b,a);
    else printf("No substring"); 
    return 0;
}

1.7.24 word length

The following statement modifies just fine [This tells us that \ (<iostream> \) unreliable]
`` `
cout >> S >>", "; // original

printf ( "% d,", s); // Now

有可能""（双引号）打成‘’（单引号）了一开始没发现
  
  所以还是$printf$吧
  

//1.7.24

include

include

include

include

using namespace std;
char a[2001];
int s=0;
int main(){
gets(a);
int l=strlen(a);
for(int i=0;i<l;i++)
{
if(a[i]!=' ') s++;
else if(a[i]==' '&&a[i-1]!=' ')
{
printf("%d,",s);
s=0;
}
}
cout<<s;
return 0;
}

# 1.7.29 $ISBN$号码
             
忘了$k==10$的特判‘$X$’

//1.7.29

include

include

include

include

using namespace std;
char a[15];
int b[15];
int k;
int main( ){
scanf("%s",a+1);
int j=1;
for(int i=1;i<=11;i++){
if(i==2||i==6) continue;
else {
k+=(a[i]-'0')*j;
j++;
}
}
//printf("%d\n",k);
k%=11;
if(k==10) k='X'-'0';
if(k==a[13]-'0') printf("Right");
else{
a[13]=k+'0';
printf("%s",a+1);
}
return 0;
}

重点！！！
                           

=. 11% K;
IF (K == 10) = K 'X-' - '0'; // on operation between characters
if (k == a [13] - '0') printf ( "Right") ;
the else {
A [13 is] = K + '0';
the printf ( "% S", A +. 1);
}

# 1.7.32 行程缩写
                           
在程序内开数组，要memset要不然输出的数很怪

//1.7.32

include

include

include

include

include

using namespace std;
int main( ){
char s[1006];
scanf("%s",s+1);
int a[1010];
memset(a,0,sizeof(a));//!!
int len1;
len1=strlen(s+1);
for(int i=1;i<=len1;i++){
if(s[i]>='a'&&s[i]<='z') s[i]=s[i]-'a'+'A';
}
//printf("%s",s+1);
for(int i=1;i<=len1;i++){
if(s[i]==s[i+1]) a[s[i]-'A']++;
else {
printf("(%c,%d)",s[i],a[s[i]-'A']+1);
a[s[i]-'A']=0;
}
}
return 0;
}

             
# 1.7.35 字符串的展开

//1.7.35

include

include

include

include

include

using namespace std;
bool iszimu(char a){
if(a>='a'&&a<='z') return true;
return 0;
}
void solve(string &s,int p1,int p2,int p3){
if(p1==1)
{
for(int i=0;i<s.length();i++)
{
if(s[i]=='-')
{
if(iszimu(s[i-1])&&iszimu(s[i+1])&&s[i+1]>s[i-1]+1)
{
if(p3==1)
{
for(char temp = s[i-1]+1;temp<s[i+1];temp++)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
else
{
for(char temp = s[i+1]-1;temp>=s[i-1]+1;temp--)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
continue;
}
else if(isdigit(s[i-1])&&isdigit(s[i+1])&&s[i+1]>s[i-1]+1)
{
if(p3==1)
{
for(char temp = s[i-1]+1;temp<s[i+1];temp++)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
else
{
for(char temp = s[i+1]-1;temp>=s[i-1]+1;temp--)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
continue;
}
else if(s[i-1]>=s[i+1] ) cout<<'-';
else if(s[i-1]+1==s[i+1]);
else cout<<'-';//刚开始没有这个，如果s-2这种就会出错（少一个-）
}
else
cout<<s[i];
}
cout<<endl;
}
else if(p1==2)
{
for(int i=0;i<s.length();i++)
{
if(s[i]=='-')
{
if(iszimu(s[i-1])&&iszimu(s[i+1])&&s[i+1]>s[i-1]+1)
{
if(p3==1)
{
for(char temp = s[i-1]+1-('a'-'A');temp<s[i+1]-('a'-'A');temp++)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
else
{
for(char temp = s[i+1]-1-('a'-'A');temp>=s[i-1]+1-('a'-'A');temp--)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
continue;
}
else if(isdigit(s[i-1])&&isdigit(s[i+1])&&s[i+1]>s[i-1]+1)
{
if(p3==1)
{
for(char temp = s[i-1]+1;temp<s[i+1];temp++)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
else
{
for(char temp = s[i+1]-1;temp>=s[i-1]+1;temp--)
{
for(int j=0;j<p2;j++) cout<<temp;
}
}
continue;
}
else if(s[i-1]>=s[i+1] ) cout<<'-';
else if(s[i-1]+1==s[i+1]);
else cout<<'-';
}
else
cout<<s[i];
}
printf("\n");
}
else
{
for(int i=0;i<s.length();i++)
{
if(s[i]=='-')
{
if(iszimu(s[i-1])&&iszimu(s[i+1])&&s[i+1]>s[i-1]+1)
{
for(char temp = s[i-1]+1;temp<s[i+1];temp++)
{
for(int j=0;j<p2;j++) cout<<'';
}
continue;
}
else if(isdigit(s[i-1])&&isdigit(s[i+1])&&s[i+1]>s[i-1]+1)
{
for(char temp = s[i-1]+1;temp<s[i+1];temp++)
{
for(int j=0;j<p2;j++) cout<<'';
}
continue;
}
else if(s[i-1]>=s[i+1] ) cout<<'-';
else if(s[i-1]+1==s[i+1]);
else cout<<'-';
}
else
cout<<s[i];
}
cout<<endl;
}
}
int main(){
int p1,p2,p3;
string s;
while(cin>>p1>>p2>>p3)
{
cin>>s;
solve(s,p1,p2,p3);
}
return 0;
}

# 最小公倍数·最大公因数
### 辗转相除法
重要定理：如果$a,b$的最大公因数和最小公倍数是$m,n$，那么$ab=mn$
  

include

include

include

using namespace std;
int gcd(int x,int y){
int r=x%y;
while(r!=0){
x=y;
y=r;
r=x%y;
}
return y;
}
int lcm(int x,int y){
return x*y/gcd(x,y);
}
int main( ){
int a,b;
printf("Please input two numbers a and b");
scanf("%d%d",&a,&b);
printf("a=%d,b=%d\n",a,b);
printf("gcd(%d,%d)=%d,lcm(%d,%d)=%d",a,b,gcd(a,b),a,b,lcm(a,b));
}
↑辗转相除法求最大公因数和最小公倍数 ### 以1.7.30字符环为例cpp
//1.7.30 字符环

include

include

include

include

using namespace std;
int gcd(int x,int y){
int r=x%y;
while(r!=0){
x=y;
y=r;
r=x%y;
}
return y;
}
int lcm(int x,int y){
return x*y/gcd(x,y);
}
int main( ){
char s1[260],s2[260];
scanf("%s%s",s1+1,s2+1);
int len1,len2;
len1=strlen(s1);len2=strlen(s2);

if(len2>len1){
    swap(len1,len2);
    swap(s1,s2);
}
int lengcd;
lengcd=gcd(len1,len2);
for(int i=1;i<=len1;i++){
    for(int j=1;j<=len1+len2,j++){
        int k=j%len2;
        while(s1[k])
    }
}

}
```

# Count the number of words
`` `cpp

include

include

using namespace std;
char s[105];
int nw=1,count=0;
//way1
int main( ){
gets(s+1);
s[0]=' ';
int le=strlen(s+1);
//printf("%d \n",le);
for(int i=0;i<=le;i++){
if(s[i]==' '){
nw=1;
}
else{
if(nw==1){
nw=0;
count++;
}
}
// printf("%d %d %c\n",i,count,s[i]);
}
printf("%d",count);
return 0;
}
void way2( ){
gets(s);
for(int i=0;i<=101;i++){
if(s[i]=='\0'){
break;
}
if(s[i]==' '){
nw=1;
}
else{
if(nw==1){
nw=0;
count++;
}
}
//printf("%d %d %c\n",i,count,s[i]);
}
printf("%d",count);
return ;
}

include

include

include

STD namespace the using;
char SM [105];
char STR [105];
int CNT = 0;
void way3 () {
Stack <int> SP;
the gets (SM);
int Le = strlen (SM);
sp.push (0 ); // advance is pressed into the space
for (int I = 0; I <Le; I ++) {
IF (SM [I] = '') sp.push (SM [I] - '0');!
the else SP. push (0); // process the string
}
the while (sp.size () = 0!) {
char tmp;
tmp = sp.top ();
IF (tmp == 0) CNT ++;
sp.pop (); / / statistics
}
the printf ( "% D", CNT);
return;
}
`` `