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codechef September Challenge 2018 Division 2 …

2018-09-18 06:25:22来源：博客园阅读 ()

比赛链接

上紫啦hhh，好开心

每个题里面都有中文翻译，我就不说题意了。。

A

直接模拟即可

#include<cstdio>
#include<algorithm>
#define int long long 
using namespace std;
const int MAXN = 1e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int T;
int a[MAXN], pos, Q, N;
main() {
    T = read();
    while(T--) {
        N = read(); pos = read(); Q = read();
        for(int i = 1; i <= N; i++) a[i] = i;
        while(Q--) {
            int x = read(), y = read();
            swap(a[x], a[y]);
        }
        for(int i = 1; i <= N; i++)
            if(pos == a[i]) {
                printf("%d\n", i); break;
            }
    }
}

B

这题我交了五遍没过,捂脸，然而并不知道自己哪里错了

跟lyq说了说还被婊了一顿。。。。最后换了个写法A了。。

直接大力特判就行，至于哪个share什么玩意儿就让N-1，M-1，然后判是否可行

#include<cstdio>
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, X, Y;
int pd(int N, int M) {
    /*if(N == 0 && M == 0) return 1;
    else if((N < X) || (M < Y)) return 0;
    else if(N % X == 0 && M % Y == 0) return 1;
    else return 0;*/
    if(N < 0 || M < 0) return 0;
    return N % X == 0 && M % Y == 0;
}
int main() {
    int T;
    scanf("%d", &T);
    while(T--) {
        N = read(), M = read(), X = read(), Y = read();
        N--; M--;
        if(pd(N, M)) {puts("Chefirnemo"); continue;}
        N--; M--;
        if(pd(N, M)) {puts("Chefirnemo"); continue;}
        puts("Pofik");
    }
    return 0;
}

C

这题就有点厉害了！

结论：哥德巴赫猜想在信息学奥赛的范围内是成立的！

也就是说，任意一个$>=4$的偶数都是合法的。

然后把奇数和$0$判掉，维护出$0, 1$的个数即可

#include<cstdio>
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int T;
int N;
int a[MAXN];
LL tim[MAXN];
LL calc(LL x) {
    return x * (x - 1) / 2;
}
int main() {
  //  printf("%d", 2 ^ 4);
    T = read();
    while(T--) {
        N = read();
        for(int i = 1; i <= N;i ++) a[i] = read(), tim[a[i]]++;
        LL ans = 0, ze = 0, on = 0;
        /*for(int i = 1; i <= N; i++) {
            for(int j = i + 1; j <= N; j++) {
                int x = (a[i] ^ a[j]);
                if((!(x & 1)) && (x != 2) && (x != 0)) ans++;
            }
        }*/

        for(int i = 1; i <= N; i++)
            if(a[i] & 1) on++;
            else ze++;
        ans = calc(ze) + calc(on);
        //printf("%d\n", ans);
        LL a1 = 0;
        for(int i = 1; i <= N; i++)
            a1 += tim[a[i] ^ 2];
        ans -= a1 / 2;
        for(int i = 1; i <= N; i++) {
            ans -= (tim[a[i]] * (tim[a[i]] - 1)) / 2, tim[a[i]] = 0;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

D

爆搜打出前$8$的表，不难找到规律：

最小的一定是$n, 1, 2, 3, \dots n - 1$

最大的是：$2, 3, 4, 5, n / 2, 1, 2 + n / 2, 3 + n / 2, 4 + n /2 ...$

奇数的话还要特殊考虑一下倒数第二位

#include<cstdio>
#include<map>
#include<iostream>
#define ull unsigned long long 
#define LL long long 
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e9 + 7;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N;
int ans[MAXN], num;
int main() {
//    freopen("a.out", "w", stdout);
    N = read();
    
    for(int i = 2; i <= N / 2; i++) ans[++num] = i; 
    ans[++num] = 1;
    for(int i = 2; i <= N / 2; i++) ans[++num] = N / 2 + i;
    ans[++num] = 1 + N / 2;
    for(int i = 1; i <= num - 1; i++)
        printf("%d ", ans[i]);
    if(N & 1) printf("%d ", N);
    printf("%d\n", ans[num]);
    printf("%d ", N);
    for(int i = 1; i <= N - 1; i++)
        printf("%d ",i);
    return 0;
}

E

首先考虑暴力怎么打，我们把给出的初始行列的01取反，这样$0$的时候对应的是必胜态，$1$对应的是必败态。

然后按博弈论的定义推，$(i, j)$若是必胜态，那么至少有$(i - 1, j)$是必败态或者 $(i, j - 1)$是必败态。

然后暴力枚举一遍就行了，复杂度$O(NM)$

接下来的操作就比较神仙了，，打表或直接观察式子可得，若第$(i, j)$个点是必败点，那么它所在的对角线往后的点，都是必败点！

这样我们就把状态降到了$2 * M$

那最开始的必败点怎么求呢？

直觉告诉我们稍加证明不难得到，这种点一定是出现在前两行或者前两列。

然后就做完了。

暴力推前两行前两列即可。

保险起见我推了三行三列。

#include<cstdio>
#include<cstring>
#include<vector>
#define LL long long 
using namespace std;
const int MAXN = 1e5 + 10;
inline LL read() {
    char c = getchar(); LL x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int T;
int a[4][MAXN], b[MAXN][4], N, M, f[MAXN], g[MAXN];
char A[MAXN], B[MAXN];
int main() { 
//-    freopen("a.out", "w", stdout);
    int T = read();
    while(T--) {
        scanf("%s", A + 1);
        scanf("%s", B + 1);
        M = strlen(A + 1);
        N = strlen(B + 1);
        for(int i = 1; i <= M; i++) a[0][i] = (A[i] == '1' ? 0 : 1);
        for(int i = 1; i <= 3; i++) a[i][0] = (B[i] == '1' ? 0 : 1);
        
        for(int i = 1; i <= 3; i++) b[0][i] = (A[i] == '1' ? 0 : 1);
        for(int i = 1; i <= N; i++) b[i][0] = (B[i] == '1' ? 0 : 1);
        
        memset(f, 0x7f, sizeof(f));
        memset(g, 0x7f, sizeof(g));
        
        for(int i = 1; i <= 3; i++) {
            for(int j = 1; j <= M; j++) {
                if(a[i - 1][j] == 1 || a[i][j - 1] == 1) a[i][j] = 0;//能到达必败节点的 
                else a[i][j] = 1;
                if(a[i][j] == 1) f[j - i + 1] = min(f[j - i + 1], i);
            }
        }
        
        for(int i = 1; i <= N; i++) {
            for(int j = 1; j <= 3; j++) {
                if(b[i - 1][j] == 1 || b[i][j - 1] == 1) b[i][j] = 0;
                else b[i][j] = 1;
                if(b[i][j] == 1) g[i - j + 1] = min(g[i - j + 1], j);
            }
        }
        vector<int> ans;
        int Q = read();
        while(Q--) {
            int x = read(), y = read();
            int dy = y - x + 1,
                dx = x - y + 1;
            if(dy >= 1) {
                if(x >= f[dy]) ans.push_back(0);
                else ans.push_back(1); 
            } else {
                if(y >= g[dx]) ans.push_back(0);
                else ans.push_back(1);
            }
        }
        for(int i = 0; i < ans.size(); i++) printf("%d", ans[i]);        
        puts("");
    }

    return 0;
}