【AtCoder】Yahoo Programming Contest 2019

其他 2019-02-10 14:08:01 阅读次数: 0

A - Anti-Adjacency

K <= (N + 1) / 2

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 40005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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 N,K;
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(N);read(K);
    if(K <= (N + 1) / 2) puts("YES");
    else puts("NO");
}

B - Path

点度都不超过3

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 40005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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 deg[5];
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    int a,b;
    for(int i = 1 ; i <= 3 ; ++i) {
        read(a);read(b);
        ++deg[a];++deg[b];
    }
    for(int i = 1 ; i <= 4 ; ++i) {
        if(deg[i] >= 3) {puts("NO");return 0;}
    }
    puts("YES");
}

C - When I hit my pocket...

先给自己加到A块饼干
然后判断一下B - A是否大于2，然后就用两次不断的换B - A
否则就直接一直拍得一次饼干就行

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 40005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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);
}
int64 K,A,B;
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    read(K);read(A);read(B);
    int64 now = 1;
    if(now < A) {
        int64 t = min(A - now,K);
        now += t;
        K -= t;
    }
    if(B - A > 2) {
        now += K / 2 * (B - A);
        K %= 2;
    }
    now += K;
    out(now);enter;
}

D - Ears

DEF我做题顺序是反的，D题我做的有点智障
FE做完之后一开榜19，D做完掉到45
大家AK水平那么高= =
事实上就是一个有点智障的dp
空 偶 奇 偶 空
每个段都可以为0，如果一个数在空段代价是它的值，奇数在偶段代价为1，偶数在奇数段代价为1，0在偶数段代价为2

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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 L;
int64 dp[MAXN][6],A[MAXN];
void Solve() {
    read(L);
    for(int i = 1 ; i <= L ; ++i) {read(A[i]);}
    for(int i = 1 ; i <= 4 ; ++i) dp[0][i] = 1e16;
    dp[0][0] = 0;
    for(int i = 1 ; i <= L ; ++i) {
        dp[i][0] = dp[i - 1][0] + A[i];
        int t = A[i] & 1;
        dp[i][1] = min(dp[i - 1][1],dp[i - 1][0]);

        dp[i][2] = min(min(dp[i - 1][1],dp[i - 1][0]),dp[i - 1][2]);
        if(A[i] % 2 == 0) dp[i][2]++;
        dp[i][3] = min(dp[i - 1][2],dp[i - 1][3]);
        if(A[i] & 1) {dp[i][1]++;dp[i][3]++;}
        if(A[i] == 0) {dp[i][1] += 2;dp[i][3] += 2;}
        dp[i][4] = min(min(dp[i - 1][3],dp[i - 1][2]),min(dp[i - 1][1],dp[i - 1][4]));
        dp[i][4] += A[i];
    }
    int64 res = dp[L][0];
    for(int i = 1 ; i <= 4 ; ++i) res = min(res,dp[L][i]);
    out(res);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Odd Subrectangles

如果选的行数确定了，那么选某一列的奇偶性也会确定，如果奇数有a列，偶数有b列，\(a + b = M\)
那么方案数是选a中奇数个，b随便选
方案数是\(2^{a - 1}\cdot 2^{b}\)，发现这个就是\(2^{M - 1}\)
所以只要存在a即可，即选的行数的数异或起来值不为0，转而求异或为0的方案，直接上线性基即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 40005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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);
}
const int MOD = 998244353;
int N,M;
int a[305][305];
int b[305][305],cnt;
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
        if(c & 1) res = mul(res,t);
        t = mul(t,t);
        c >>= 1;
    }
    return res;
}
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 1 ; j <= M ; ++j) {
            read(a[i][j]);
        }
    }
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = 1 ; j <= M ; ++j) {
            if(a[i][j]) {
                if(!b[j][j]) {
                    for(int k = 1 ; k <= M ; ++k) b[j][k] = a[i][k];
                    ++cnt;
                    break;
                }
                else {
                    for(int k = 1 ; k <= M ; ++k) a[i][k] ^= b[j][k];
                }
            }
        }
    }
    int ans = inc(fpow(2,N),MOD - fpow(2,N - cnt));
    ans = mul(ans,fpow(2,M - 1));
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Pass

直接转二维平面，然后第i步能走到的点\((a,b)\)a个红球，b个蓝球，要满足a小于等于前i个人红球总和，b小于等于前i个人蓝球总和

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 40005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    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);
}
const int MOD = 998244353;
char s[2005];
int sum[2005][2],N;
int dp[4005][4005];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a * b % MOD;
}
void update(int &x,int y) {
    x = inc(x,y);
}
void Solve() {
    scanf("%s",s + 1);
    N = strlen(s + 1);
    for(int i = 1 ; i <= N ; ++i) {
        sum[i][0] = sum[i - 1][0];
        sum[i][1] = sum[i - 1][1];
        if(s[i] == '0') sum[i][0] += 2;
        if(s[i] == '1') {sum[i][0]++;sum[i][1]++;}
        if(s[i] == '2') sum[i][1] += 2;
    }
    dp[0][0] = 1;
    for(int i = 1 ; i <= 2 * N ; ++i) {
        int t = min(i,N);
        for(int j = 0 ; j <= i ; ++j) {
            int a = j,b = i - j;
            if(a <= sum[t][0] && b <= sum[t][1]) {
                if(a) update(dp[a][b],dp[a - 1][b]);
                if(b) update(dp[a][b],dp[a][b - 1]);
            }
        }
    }
    out(dp[sum[N][0]][sum[N][1]]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

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