AtCoder Grand Contest 019 B

题意：
可以选择一个子串翻转一次 能得到多少个不同的串

做法：
考虑一个串的首尾是相同的字母
假设它的长度为n
那么翻转[1,n] 等价于 翻转[2,n-1]
所以忽略掉翻转[1,n]的情况

问题就转化为求有多少对相同的字母
总共可翻转的选择减去这些情况就是答案

#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <queue>#include <cstdio>#include <map>#include <set>#include <utility>#include <stack>#include <cstring>#include <cmath>#include <vector>#include <ctime>#include <bitset>using namespace std;#define pb push_back#define sd(n) scanf("%d",&n)#define sdd(n,m) scanf("%d%d",&n,&m)#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)#define sld(n) scanf("%lld",&n)#define sldd(n,m) scanf("%lld%lld",&n,&m)#define slddd(n,m,k) scanf("%lld%lld%lld",&n,&m,&k)#define sf(n) scanf("%lf",&n)#define sff(n,m) scanf("%lf%lf",&n,&m)#define sfff(n,m,k) scanf("%lf%lf%lf",&n,&m,&k)#define ss(str) scanf("%s",str)#define ans() printf("%d",ans)#define ansn() printf("%d\n",ans)#define anss() printf("%d ",ans)#define lans() printf("%lld",ans)#define lanss() printf("%lld ",ans)#define lansn() printf("%lld\n",ans)#define fansn() printf("%.5f\n",ans)#define r0(i,n) for(int i=0;i<(n);++i)#define r1(i,e) for(int i=1;i<=e;++i)#define rn(i,e) for(int i=e;i>=1;--i)#define rsz(i,v) for(int i=0;i<(int)v.size();++i)#define szz(x) ((int)x.size())#define mst(abc,bca) memset(abc,bca,sizeof abc)#define lowbit(a) (a&(-a))#define all(a) a.begin(),a.end()#define pii pair<int,int>#define pli pair<ll,int>#define mp make_pair#define lrt rt<<1#define rrt rt<<1|1#define X first#define Y second#define PI (acos(-1.0))#define sqr(a) ((a)*(a))typedef long long ll;typedef unsigned long long ull;const int mod = 1000000000+7;const double eps=1e-9;const int inf=0x3f3f3f3f;const ll infl = 10000000000000000;const int maxn=  200000+10;const int maxm = 10000+10;int in(int &ret){    char c;    int sgn ;    if(c=getchar(),c==EOF)return -1;    while(c!='-'&&(c<'0'||c>'9'))c=getchar();    sgn = (c=='-')?-1:1;    ret = (c=='-')?0:(c-'0');    while(c=getchar(),c>='0'&&c<='9')ret = ret*10+(c-'0');    ret *=sgn;    return 1;}char s[maxn];int cnt[30];int main(){#ifdef LOCAL    freopen("input.txt","r",stdin);//    freopen("output.txt","w",stdout);#endif // LOCAL//    freopen("kingdom.in","r",stdin);//    freopen("kingdom.out","w",stdout);    ss(s+1);    int n = strlen(s+1);    for(int i=1;i<=n;++i)    {        int idx = s[i]-'a';        cnt[idx]++;    }    ll ans = 1LL*(n-1)*n/2 +1 ;    for(int i=0;i<26;++i)    {        int a = cnt[i];        ans -= 1LL*(a-1)*a/2;    }    lansn();    return 0;}