ZOJ 3490 String Successor
2013-08-01 21:41
239 查看
题意:给出一串可见字符和一个整数N,每次对这串字符的最右边的某个字符+1,且这个字符必须是数字或大小写字母。如果这个字符+1后超过了这个字符的范围就向右产生一个进位,而这个字符变为字符范围的最小字符,即若字符为9,+1后变为0,然后产生进位传给左边的字母或数字符号,依次下去,如果有必要要在字符的最左边插入一个字符。(题目叙述挺搓的。)
分析:递归模拟。
Code:
#include <algorithm>
#include <iostream>
#include <memory.h>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#define eps 1e-8
#define LL long long
#define pb push_back
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=100005;
string str;
bool nonalnum(int id){
for(int i=0;i<=id;i++){
if(isalnum(str[i])) return false;
}
return true;
}
void add(int last){
for(int i=last;i>=0;i--){
if(isalnum(str[i])){
if(str[i]=='9'){
str[i]='0';
if(i==0||nonalnum(i-1)) str.insert(i,"1");
else add(i-1);
}
else if(str[i]=='z'){
str[i]='a';
if(i==0||nonalnum(i-1)) str.insert(i,"a");
else add(i-1);
}
else if(str[i]=='Z'){
str[i]='A';
if(i==0||nonalnum(i-1)) str.insert(i,"A");
else add(i-1);
}
else str[i]++;
return ;
}
}
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--){
cin>>str>>n;
for(int i=0;i<n;i++){
if(nonalnum(str.size()-1)) str[str.size()-1]++;
else add(str.size()-1);
cout<<str<<endl;
}
cout<<endl;
}
return 0;
}
分析:递归模拟。
Code:
#include <algorithm>
#include <iostream>
#include <memory.h>
#include <cstring>
#include <cstdlib>
#include <string>
#include <cstdio>
#include <vector>
#include <cctype>
#include <queue>
#include <cmath>
#include <map>
#include <set>
#define eps 1e-8
#define LL long long
#define pb push_back
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int inf=0x3f3f3f3f;
const int maxn=100005;
string str;
bool nonalnum(int id){
for(int i=0;i<=id;i++){
if(isalnum(str[i])) return false;
}
return true;
}
void add(int last){
for(int i=last;i>=0;i--){
if(isalnum(str[i])){
if(str[i]=='9'){
str[i]='0';
if(i==0||nonalnum(i-1)) str.insert(i,"1");
else add(i-1);
}
else if(str[i]=='z'){
str[i]='a';
if(i==0||nonalnum(i-1)) str.insert(i,"a");
else add(i-1);
}
else if(str[i]=='Z'){
str[i]='A';
if(i==0||nonalnum(i-1)) str.insert(i,"A");
else add(i-1);
}
else str[i]++;
return ;
}
}
}
int main()
{
int T,n;
scanf("%d",&T);
while(T--){
cin>>str>>n;
for(int i=0;i<n;i++){
if(nonalnum(str.size()-1)) str[str.size()-1]++;
else add(str.size()-1);
cout<<str<<endl;
}
cout<<endl;
}
return 0;
}
相关文章推荐
- zoj 3490 String Successor
- zoj 3490
- ZOJ 3490 String Successor(模拟)
- ZOJ - 3490 String Successor 递归模拟
- ZOJ 3490 String Successor
- ZOJ 3490 String Successor
- ZOJ 3490 String Successor 字符串处理
- ZOJ 3490 String Successor(模拟啊 )
- ZOJ 3490 String Successor(字符串处理)
- zoj 3490 String Successor 模拟
- zoj 3490 string successor
- ZOJ- 3490 String Successor 模拟
- ZOJ 3490 String Successor(模拟)
- ZOJ 3490 String Successor//字符串处理
- ZOJ 3490 String Successor
- ZOJ Problem Set - 3490 String Successor
- ZOJ 3490 String Successor 字符串处理
- ZOJ 3490 String Successor(模拟)
- String Successor(ZOJ - 3490 )
- ZOJ-3490-String Successor【8th浙江省赛】【模拟】