题意:给定一个字符串,让你把它的一个子串字符都减1,使得总字符串字典序最小。
析:由于这个题是必须要有一个字串,所以你就要注意这个只有一个字符a的情况,其他的就从开始减 1,如果碰到a了就不减了,如果到最后一位了还没开始减,
就减最后一位。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 5;
const int mod = 1e8;
const char *mark = "+-*";
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, -1, 1, 1, -1};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
char s[maxn]; int main(){
while(scanf("%s", s) == 1){
int len = strlen(s);
int ok = false;
for(int i = 0; i < len; ++i){
if(s[i] =='a' && ok) break;
if(i == len-1 && !ok){
s[i] = (s[i]-'a'+25) % 26 + 'a';
break;
}
if(s[i] != 'a'){
ok = true;
--s[i];
}
}
printf("%s\n", s);
}
return 0;
}