标签：syn empty ack input substring stream namespace nbsp you
You are given a non-empty string s consisting of lowercase English letters. You have to pick exactly one non-empty substring of s and shift all its letters ‘z‘ 
‘y‘ ‘x‘ 
‘b‘ ‘a‘ ‘z‘. In other words, each character is replaced with the previous character of English alphabet and ‘a‘ is replaced with ‘z‘.
What is the lexicographically minimum string that can be obtained from s by performing this shift exactly once?
The only line of the input contains the string s (1?≤?|s|?≤?100?000) consisting of lowercase English letters.
Print the lexicographically minimum string that can be obtained from s by shifting letters of exactly one non-empty substring.
codeforces
bncdenqbdr
abacaba
aaacaba
String s is lexicographically smaller than some other string t of the same length if there exists some 1?≤?i?≤?|s|, such that s1?=?t1,?s2?=?t2,?...,?si?-?1?=?ti?-?1, and si?<?ti.
唯一的坑点就是aaaaaa...时，最后一个要该为z；
因为题目说明不能为空子串；
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == ‘-‘) f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
char ch[30];
map<char, char>Map;
int main()
{
//ios::sync_with_stdio(0);
for (int i = 1; i <= 26; i++)ch[i] = ‘a‘ + i - 1;
ch[0] = ‘z‘;
for (int i = 1; i <= 26; i++)Map[ch[i]] = ch[i - 1];
string s; cin >> s; int i;
int len = s.length(); int pos = 0; int fg = 0;
int cnt = 0;
for (int i = 0; i < len; i++) {
if (s[i] == ‘a‘)continue;
else { fg = 1; break; }
}
if (fg == 0) {
for (int i = 0; i < len - 1; i++)cout << s[i];
cout << ‘z‘ << endl;
return 0;
}
fg = 0;
for (i = 0; i < len; i++) {
if (Map[s[i]] < s[i]) {
cout << Map[s[i]]; fg = 1;
cnt++;
}
else {
if (fg == 0) {
cout << s[i]; continue;
}
else if (fg) {
pos = i;
break;
}
}
}
if (i < len) {
if (fg)for (int i = pos; i < len; i++)cout << s[i];
}
cout << endl;
return 0;
}
CF708A Letters Cyclic Shift 模拟
标签：syn empty ack input substring stream namespace nbsp you
原文地址：https://www.cnblogs.com/zxyqzy/p/10127187.html
