【CF】283D Tennis Game

枚举t加二分判断当前t是否可行，同时求出s。
注意不能说|a[n]| <= |3-a[n]|就证明无解，开始就是wa在这儿了。
可以简单想象成每当a[n]赢的时候，两人都打的难解难分（仅多赢一轮）；而每当a[n]输的时候，一轮都没赢。
在这个前提下，显然存在|a[n]| <= |3-a[n]|。

 /* 283D */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int maxn = 1e5+;
 int a[maxn];
 int cnt[][maxn];
 int n;
 int id;

 int calc(int t) {
     int i, j, p, q;
     int c[];
     int b[];

     b[] = b[] = ;
     c[] = c[] = ;
     p = ;
     while () {
         i = lower_bound(cnt[]+p, cnt[]++n, c[]+t) - (cnt[]);
         j = lower_bound(cnt[]+p, cnt[]++n, c[]+t) - (cnt[]);
         if (cnt[][i]-c[]!=t && cnt[][j]-c[]!=t)
             return ;
         if (i < j) {
             // 1 win
             q = ;
             ++b[];
             p = i;
         } else {
             q = ;
             ++b[];
             p = j;
         }
         c[] = cnt[][p];
         c[] = cnt[][p];
         if (p >= n)
             break;
     }

     if (p != n)
         return ;

     int id_ =  - id;
     if (b[id_]>=b[id] || q!=id)
         return ;
     return b[id];
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     int c[];

     scanf("%d", &n);
     c[] = c[] = ;
     rep(i, , n+) {
         scanf("%d", &a[i]);
         ++c[a[i]];
         cnt[a[i]][i] = cnt[a[i]][i-] + ;
         cnt[-a[i]][i] = cnt[-a[i]][i-];
     }
     cnt[][n+] = cnt[][n+] = INT_MAX;

     id = a[n];
     int an = c[id], bn = c[]+c[]-an;

     // if (an <= bn) {
         // puts("0");
         // return 0;
     // }

     int i, j;
     vpii ans;

     for (i=; i<=n; ++i) {
         j = calc(i);
         if (j)
             ans.pb(mp(j, i));
     }

     sort(all(ans));

     n = SZ(ans);
     printf("%d\n", n);
     rep(i, , n) {
         printf("%d %d\n", ans[i].fir, ans[i].sec);
     }

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }