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]|。
1 /* 283D */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23 24 #define sti set<int> 25 #define stpii set<pair<int, int> > 26 #define mpii map<int,int> 27 #define vi vector<int> 28 #define pii pair<int,int> 29 #define vpii vector<pair<int,int> > 30 #define rep(i, a, n) for (int i=a;i<n;++i) 31 #define per(i, a, n) for (int i=n-1;i>=a;--i) 32 #define clr clear 33 #define pb push_back 34 #define mp make_pair 35 #define fir first 36 #define sec second 37 #define all(x) (x).begin(),(x).end() 38 #define SZ(x) ((int)(x).size()) 39 #define lson l, mid, rt<<1 40 #define rson mid+1, r, rt<<1|1 41 42 const int maxn = 1e5+5; 43 int a[maxn]; 44 int cnt[3][maxn]; 45 int n; 46 int id; 47 48 int calc(int t) { 49 int i, j, p, q; 50 int c[3]; 51 int b[3]; 52 53 b[1] = b[2] = 0; 54 c[1] = c[2] = 0; 55 p = 0; 56 while (1) { 57 i = lower_bound(cnt[1]+p, cnt[1]+1+n, c[1]+t) - (cnt[1]); 58 j = lower_bound(cnt[2]+p, cnt[2]+1+n, c[2]+t) - (cnt[2]); 59 if (cnt[1][i]-c[1]!=t && cnt[2][j]-c[2]!=t) 60 return 0; 61 if (i < j) { 62 // 1 win 63 q = 1; 64 ++b[1]; 65 p = i; 66 } else { 67 q = 2; 68 ++b[2]; 69 p = j; 70 } 71 c[1] = cnt[1][p]; 72 c[2] = cnt[2][p]; 73 if (p >= n) 74 break; 75 } 76 77 if (p != n) 78 return 0; 79 80 int id_ = 3 - id; 81 if (b[id_]>=b[id] || q!=id) 82 return 0; 83 return b[id]; 84 } 85 86 int main() { 87 ios::sync_with_stdio(false); 88 #ifndef ONLINE_JUDGE 89 freopen("data.in", "r", stdin); 90 freopen("data.out", "w", stdout); 91 #endif 92 93 int c[3]; 94 95 scanf("%d", &n); 96 c[1] = c[2] = 0; 97 rep(i, 1, n+1) { 98 scanf("%d", &a[i]); 99 ++c[a[i]]; 100 cnt[a[i]][i] = cnt[a[i]][i-1] + 1; 101 cnt[3-a[i]][i] = cnt[3-a[i]][i-1]; 102 } 103 cnt[1][n+1] = cnt[2][n+1] = INT_MAX; 104 105 id = a[n]; 106 int an = c[id], bn = c[1]+c[2]-an; 107 108 // if (an <= bn) { 109 // puts("0"); 110 // return 0; 111 // } 112 113 int i, j; 114 vpii ans; 115 116 for (i=1; i<=n; ++i) { 117 j = calc(i); 118 if (j) 119 ans.pb(mp(j, i)); 120 } 121 122 sort(all(ans)); 123 124 n = SZ(ans); 125 printf("%d ", n); 126 rep(i, 0, n) { 127 printf("%d %d ", ans[i].fir, ans[i].sec); 128 } 129 130 #ifndef ONLINE_JUDGE 131 printf("time = %d. ", (int)clock()); 132 #endif 133 134 return 0; 135 }