Codeforces 451D Count Good Substrings (组合数)

题意

给出一个只有a和b的串求有多少个子串满足删去连续的重复字母之后是一个回文串，输出分为奇数长度和偶数长度的。

思路

只有两个字母这是个很重要的条件，这样的话一个子串只需要头和尾相同就可以删成一个回文串了。
那就分别讨论一下就可以了：
奇数长度的=C(2oddnumofa)+C(2oddnumofb)+C(2evennumofa)+C(2evennumofb)
偶数长度的=oddnumofa∗evennumofa+oddnumofb∗evennumofa

代码

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <stack>#include <set>#include <map>#include <list>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;#define LL long long#define lowbit(x) ((x)&(-x))#define lson l, mid, rt << 1#define rson mid + 1, r, rt << 1|1#define MP(a, b) make_pair(a, b)const int INF = 0x3f3f3f3f;const int MOD = 1000000007;const int maxn = 1e5 + 10;const double eps = 1e-8;const double PI = acos(-1.0);typedef pair<int, int> pii;char s[100010];LL cal(LL x){    return x * (x - 1) / 2;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    scanf("%s", s + 1);    int len = strlen(s + 1);    LL odda = 0, oddb = 0;    LL evena = 0, evenb = 0;    for (int i = 1; i <= len; i++)    {        if (s[i] == 'a')            if (i & 1) odda++;            else evena++;        else            if (i & 1) oddb++;            else evenb++;    }    LL ans1 = cal(odda) + cal(oddb) + cal(evena) + cal(evenb) + len;    LL ans2 = odda * evena + oddb * evenb;    printf("%lld %lld\n", ans2, ans1);    return 0;}