Submission #728090

#	Time	Username	Problem	Language	Result	Execution time	Memory
728090		thomas_li	Bowling (BOI15_bow)	C++17	100 / 100	54 ms	7268 KiB

bow

//#pragma GCC optimize("Ofast,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#include <bits/stdc++.h>
using namespace std;
string rk = "x-"; vector<string> spstr,rgstr;
string c1 = "xxx";
vector<string> c2,c3,c4,c5,c6,c7;
int n,a[11]; int64_t dp[11][12][22][301]; // first i frames, carry amount from previous i, carry amount from i, score after i
                                          // carry is -1 if we don't add
string s;

bool canrg[10][10];
void solve(){
    memset(dp,0,sizeof dp);
    cin >> n >> s; s = ' ' + s;
    for(int i = 1; i <= n; i++) cin >> a[i];
    auto can = [&](int i, string& t){
        assert(t.size() == 2);
        for(int j = 0; j < 2; j++){
            char c = s[i+j]; if(c == '?') continue;
            if(c != t[j]) return false;
        }
        return true;
    };
    auto can1 = [&](int i, string& t){
        assert(t.size() == 3);
        for(int j = 0; j < 3; j++){
            char c = s[i+j];
            if(c == '?') continue;
            if(c != t[j]) return false;
        }
        return true;
    };
    dp[0][0][0][0] = 1;
    auto add = [&](int64_t& x, int64_t y){
        x = (x+y);
    };
    // solve normal
    for(int i = 1; i <= n-1; i++){
        // generate good strings
        bool grk = can(2*i-1,rk);
        vector<string> gsp,grg;  
        for(auto& t : spstr) if(can(2*i-1,t)) gsp.push_back(t);
        for(auto& t : rgstr) if(can(2*i-1,t)) grg.push_back(t);
        swap(gsp,spstr);
        swap(grg,rgstr);
        int lo = 0, hi = 30*i;
        if(a[i] != -1) lo = hi = a[i];           
        for(int j = -1; j <= 10; j++) for(int k = -1; k <= 20; k++){ // previous previous, previous carries
            // strike
            // our shot needs to make j zero, k non negative
            if(grk && (j == -1 || j-10 == 0) && (k == -1 || (k-10 >= 0 && k-10 <= 10))){
                int nk = (k == -1 ? -1 : k-10);
                // we can gain some more from carries if needed 
                for(int gain = 0; gain <= 20; gain++){
                    // l >= gain+10
                    //if(l-gain-10 < 0) break;
                    for(int l = max(lo,gain+10); l <= hi; l++) add(dp[i][nk+1][gain+1][l],dp[i-1][j+1][k+1][l-gain-10]);
                }
            }     
            // spare
            if(k == -1 || k-10 == 0){
                // TODO: opt break
                for(auto& t : spstr){
                    // previous previous gets subtracted by fs, previous gets subtracted by 10
                    // both are gone after this frame, so need to become 0
                    int fs = t[0]-'0';
                    if(j != -1 && j-fs != 0) continue;
                    for(int gain = 0; gain <= 10; gain++){
                        //if(l-gain-10 < 0) break;
                        // add to previous slot to satisfy one bonus feature
                        for(int l = max(lo,gain+10); l <= hi; l++) add(dp[i][gain+1][0][l],dp[i-1][j+1][k+1][l-gain-10]);
                    }
                }
            }
            // regular
            for(auto& t : rgstr){
                // previous previous gets subtracted by fs, previous gets subtracted by sn+fs
                // both are gone after this frame, so need to become 0
                int fs = t[0]-'0',sn = t[1]-'0';
                if(j != -1 && j-fs != 0) continue;
                if(k != -1 && k-fs-sn != 0) continue;
                //if(l-fs-sn < 0) continue;
                for(int l = max(lo,fs+sn); l <= hi; l++) add(dp[i][0][0][l],dp[i-1][j+1][k+1][l-fs-sn]);
            }
        }
        swap(gsp,spstr);
        swap(grg,rgstr);
    }
    // solve last
    // preprocess equalities
    vector<string> gc2,gc3,gc4,gc5,gc6,gc7;
    for(auto& t : c2) if(can1(2*n-1,t)) gc2.push_back(t);
    for(auto& t : c3) if(can1(2*n-1,t)) gc3.push_back(t);
    for(auto& t : c4) if(can1(2*n-1,t)) gc4.push_back(t);
    for(auto& t : c5) if(can1(2*n-1,t)) gc5.push_back(t);
    for(auto& t : c6) if(can1(2*n-1,t)) gc6.push_back(t);
    for(auto& t : c7) if(can1(2*n-1,t)) gc7.push_back(t);
    bool gc1 = can1(2*n-1,c1);
    swap(gc2,c2);
    swap(gc3,c3);
    swap(gc4,c4);
    swap(gc5,c5);
    swap(gc6,c6);
    swap(gc7,c7);
    int64_t res = 0;
    int lo = 0, hi = 300;
    if(a[n] != -1) lo = hi = a[n];
    for(int j = -1; j <= 10; j++) for(int k = -1; k <= 20; k++){ // previous previous, previous carries


        if((j == -1 || j-10 == 0) && (k == -1 || k-20 == 0)){
            // xxx 
            if(gc1) for(int l = max(lo,30); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-30]);
            // xxA  
            // condition is same since bonus is only contributed in the first two shots
            for(auto& t : c2) {
                int ls = t[2]-'0'; 
                for(int l = max(lo,20+ls); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-20-ls]);
            }
        }
        if((j == -1 || j-10 == 0)){
            // xA/
            for(auto& t : c3) {
                int fs = t[1]-'0';
                if(k != -1 && k-10-fs != 0) continue; 
                for(int l = max(lo,20); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-20]);
            }
        }
        // xAB
        for(auto& t : c4){
            int fs = t[1]-'0',sn = t[2]-'0';
            if(j != -1 && j-10 != 0) continue; 
            if(k != -1 && k-10-fs != 0) continue; 
            for(int l = max(lo,10+fs+sn); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-10-fs-sn]);
        }

        if(k == -1 || k-10 == 0){
            // A/x
            for(auto& t : c5) {
                int fs = t[0]-'0'; 
                if(j != -1 && j-fs != 0) continue;
                for(int l = max(lo,20); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-20]);
            }
            // A/B
            for(auto& t : c6) if(k == -1 || k-10 == 0){
                int fs = t[0]-'0',ls = t[2]-'0'; 
                if(j != -1 && j-fs != 0) continue;
                for(int l = max(lo,10+ls); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-10-ls]);
            }
        }
        // AB-
        for(auto& t : c7){
            int fs = t[0]-'0',sn = t[1]-'0'; 
            if(j != -1 && j-fs != 0) continue;
            if(k != -1 && k-fs-sn != 0) continue;
            for(int l = max(lo,fs+sn); l <= hi; l++) add(res,dp[n-1][j+1][k+1][l-fs-sn]);
        }
    }
    swap(gc2,c2);
    swap(gc3,c3);
    swap(gc4,c4);
    swap(gc5,c5);
    swap(gc6,c6);
    swap(gc7,c7);
    cout << res << "\n";
}
int main(){
    cin.tie(0)->sync_with_stdio(0);
    for(int i = 0; i <= 9; i++){
        spstr.push_back(to_string(i) + "/");
        c2.push_back("xx"+to_string(i));
        c3.push_back("x"+to_string(i)+"/");
        for(int j = 0; i+j < 10; j++){
            rgstr.push_back(to_string(i)+to_string(j));
            c4.push_back("x"+to_string(i)+to_string(j));
            c7.push_back(to_string(i)+to_string(j)+"-");
        }
        c5.push_back(to_string(i)+"/x");
        for(int j = 0; j <= 9; j++){
            c6.push_back(to_string(i)+"/"+to_string(j));
        }
    }
    int t; cin >> t; while(t--) solve();
}
// number of strs for frame: 10 + 1 + 10
// 5726805883325784576

• Subtask 1: 16 / 16
• Subtask 2: 17 / 17
• Subtask 3: 26 / 26
• Subtask 4: 23 / 23
• Subtask 5: 18 / 18