https://codeforces.com/contest/1321/problem/C
题意:给出一个字符串,消除规则:某个字符比左边相邻或右边相邻字符大1,则该字符可以删除,比如ba,b可以删除。
问最多可以删除多少个字符。
解法:贪心从字符z开始删除,判断能否从左边找和右边找比该字符小1的字符,找到则删除该字符,并继续从头开始枚举该字符,比如cbbbbcaaca。
//#include<bits/stdc++.h> #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <string> #include <stdio.h> #include <queue> #include <stack> #include <map> #include <set> #include <string.h> #include <vector> typedef long long ll ; #define int ll #define mod 1000000007 #define gcd __gcd #define rep(i , j , n) for(int i = j ; i <= n ; i++) #define red(i , n , j) for(int i = n ; i >= j ; i--) #define ME(x , y) memset(x , y , sizeof(x)) //ll lcm(ll a , ll b){return a*b/gcd(a,b);} ll quickpow(ll a , ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;b>>=1,a=a*a%mod;}return ans;} //int euler1(int x){int ans=x;for(int i=2;i*i<=x;i++)if(x%i==0){ans-=ans/i;while(x%i==0)x/=i;}if(x>1)ans-=ans/x;return ans;} //const int N = 1e7+9; int vis[n],prime[n],phi[N];int euler2(int n){ME(vis,true);int len=1;rep(i,2,n){if(vis[i]){prime[len++]=i,phi[i]=i-1;}for(int j=1;j<len&&prime[j]*i<=n;j++){vis[i*prime[j]]=0;if(i%prime[j]==0){phi[i*prime[j]]=phi[i]*prime[j];break;}else{phi[i*prime[j]]=phi[i]*phi[prime[j]];}}}return len} #define SC scanf #define INF 0x3f3f3f3f #define PI acos(-1) #define pii pair<int,int> #define fi first #define se second #define lson l,mid,root<<1 #define rson mid+1,r,root<<1|1 using namespace std; const int N = 1e6+100; const int maxn = 1e2+9; int flag ; char c = 'z'; string a ; bool search(int i){ if(i - 1 >= 0 && a[i - 1] == c - 1){ a.erase(a.begin() + i); return true; } if(i + 1 < a.size() && a[i + 1] == c - 1){ a.erase(a.begin() + i); return true; } return false; } void solve(){ int n ; cin >> n ; cin >> a ; while(c > 'a'){ flag = 0 ; rep(i , 0 , a.size()){ if(a[i] == c){ if(search(i)){ flag = 1 ; } } } c--; if(flag) c++; } cout << n - a.size() << endl; } signed main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); //int t ; //cin >> t ; //while(t--){ solve(); //} }