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; }