题解

设\(f[p][a][b]\)表示询问了\(p\)次，答案是\(a,b\)是否会被猜出来

然后判断如果\(p = 1\)
第一个问的\(Alice\)，那么\([s,\sqrt{nm}]\)约数只有一个，\(f[p][a][b] = 1\)否则为\(0\)
如果第一个问的\(Bob\)，那么\(a + b - 2 * S <= 1\)那么\(f[p][a][b] = 1\)否则为\(0\)

剩下的按照\(p\)这次操作询问谁，且然后从可能的数对里找\(f[p - 1][i][j]\)为\(0\)的有几个，如果只有1个就猜出来了

同时还要保证，在某个人猜出来之后，另一个猜出来的人询问第\(T + 1\)回合刚好能被猜出来的数对中，也只会问出一个，那么答案就是\(a,b\)了

用记搜，跑得还是挺快的

代码

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define enter putchar('\n')
#define space putchar(' ')
#define MAXN 20005
#define eps 1e-8
#define ivorysi
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
    while(c < '0' || c > '9') {
    if(c == '-') f = -1;
    c = getchar();
    }
    while(c >= '0' && c <= '9') {
    res = res * 10 + c - '0';
    c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
    out(x / 10);
    }
    putchar('0' + x % 10);
}
int f[25][505][505],cnt[505 * 505],g[505][505];
vector<int> v[505 * 505];
int S,T,k;
char c[25];
bool dp(int p,int a,int b) {
    if(p == 0) return 0;
    if(f[p][a][b] != -1) return f[p][a][b];
    if(p == 1) {
    if(k == 1) {
        
        if(cnt[a * b] > 1) f[p][a][b] = 0;
        else f[p][a][b] = 1;
    }
    else {
        if(a + b - 2 * S <= 1) f[p][a][b] = 1;
        else f[p][a][b] = 0;
    }
    return f[p][a][b];
    }
    if(dp(p - 1,a,b)) return f[p][a][b] = 1;
    if(p % 2 == k) {
    int c = 0;
    for(auto h : v[a * b]) {
        if(!dp(p - 1,h,a * b / h)) ++c;
    }
    if(c == 1) f[p][a][b] = 1;
    else f[p][a][b] = 0;
    }
    else {
    int c = 0;
    for(int i = S ; i <= (a + b) / 2 ; ++i) {
        if(!dp(p - 1,i,a + b - i)) ++c;
    }
    if(c == 1) f[p][a][b] = 1;
    else f[p][a][b] = 0;
    }
    return f[p][a][b];
}
bool check(int a,int b) {
    if(g[a][b] != -1) return g[a][b];
    int c = 0;
    if((T + 2) % 2 == k) {
    for(auto h : v[a * b]) {
        if(dp(T + 1,h,a * b / h) && !dp(T,h,a * b / h)) ++c;
    }
    }
    else {
    for(int i = S ; i <= (a + b) / 2 ; ++i) {
        if(dp(T + 1,i,a + b - i) && !dp(T,i,a + b - i)) ++c;
    }
    }
    if(c == 1) g[a][b] = 1;
    else g[a][b] = 0;
    return g[a][b];
}
pii Solve() {
    memset(f,-1,sizeof(f));
    memset(cnt,0,sizeof(cnt));
    memset(g,-1,sizeof(g));
    for(int i = S * S ; i <= 500 * 500 ; ++i) {
    int t = sqrt(i);
    v[i].clear();
    for(int j = S ; j <= t ; ++j) {
        if(i % j == 0) {
        cnt[i]++;
        v[i].pb(j);
        }
    }
    }
    for(int s = S + S ; s <= 1000 ; ++s) {
    for(int i = S ; i <= 500 ; ++i) {
        if(i > s) break;
        for(int j = i ; j <= s - i ; ++j) {
        if(i + j > s) break;
        if(dp(T + 1,i,j) && !dp(T,i,j) && check(i,j)) {return mp(i,j);}
        }
    }
    }
}
int main() {
    for(int i = 1 ; i <= 25 ; ++i) {
    stringstream ss;
    string num;
    ss << i;
    ss >> num;
    string str1 = "guess/guess" + num + ".in";
    FILE *fin = fopen(str1.c_str(),"r");
    fscanf(fin,"%d%s%d",&S,c + 1,&T);
       
    string str2 = "answer/guess" + num + ".out";
    FILE *fout = fopen(str2.c_str(),"w");
    if(c[1] == 'A') k = 1;
    else k = 0;
    pii ans = Solve();
    fprintf(fout,"%d %d\n",ans.fi,ans.se);
    printf("OK with %d task\n",i);
    }
    return 0;
}