Codeforces Educational Round 12 655ABCDE

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Codeforces Educational Round 12
通过数： 3
A：
刚开始用公式分类讨论，后面发现暴力才是一种更优美的解法

#include <bits/stdc++.h>using namespace std;const int MAXN = 10 + 5;char str[MAXN];int main(){    int x1, y1, x2, y2;    while(scanf("%d%d%d%d", &x1, &y1, &x2, &y2) != EOF) {        scanf("%s", str);        int st = ((str[0] - '0') * 10 + (str[1] - '0')) * 60 + (str[3] - '0') * 10 + str[4] - '0';        int en = st + y1;        int ans = 0;        int now = 300;        while(now < 1440) {            if(now + y2 > st && now < en) ans++;            now += x2;        }        printf("%d\n", ans);    }    return 0;}

B：
暴力模拟

#include <bits/stdc++.h>using namespace std;const int MAXN = 100 + 5;int id[MAXN];int pos[MAXN];int n, m, l;int remov(int mark){    int res = id[mark];    for(int i = res ; i >= 2 ; i--) {        pos[i] = pos[i - 1];        id[pos[i]] = i;    }    pos[1] = mark;    id[mark] = 1;    return res;}int main(){    while(scanf("%d%d%d", &n, &m, &l) != EOF) {        for(int i = 1 ; i <= l ; i++) {            int u; scanf("%d", &u);            pos[i] = u; id[u] = i;        }        int ans = 0;        for(int i = 1 ; i <= n ; i++) {            for(int j = 1 ; j <= m ; j++) {                int u; scanf("%d", &u);                int temp = remov(u);//                printf("i = %d, j = %d, temp = %d\n", i, j, temp);//                for(int k = 1 ; k <= l ; k++) printf("%d ", pos[k]);//                puts("");//                system("pause");                ans += temp;            }        }        printf("%d\n", ans);    }    return 0;}

C：
连续相同的一段贪心的变换一下就可以

#include <bits/stdc++.h>using namespace std;const int MAXN = 2e5 + 5;char str[MAXN];int main(){    while(scanf("%s", str) != EOF) {        int len = strlen(str);        int now = 0;        for(int i = 1 ; i < len ; i++) {            if(str[i] == str[now]) continue;            else {//                printf("i = %d, now = %d\n", i, now);                for(int j = now + 1 ; j <= i - 1 ; j += 2) {                    for(int k = 0 ; k < 26 ; k++) {//                        printf("j = %d, k = %d\n", j, k);                        if(j > 0 && str[j - 1] - 'a' == k) continue;                        if(j < len - 1 && str[j + 1] - 'a' == k) continue;                        str[j] = k + 'a';                        break;                    }                }                now = i;            }        }        int i = len;        for(int j = now + 1 ; j <= i - 1 ; j += 2) {            for(int k = 0 ; k < 26 ; k++) {//                     printf("j = %d, k = %d\n", j, k);                if(j > 0 && str[j - 1] - 'a' == k) continue;                if(j < len - 1 && str[j + 1] - 'a' == k) continue;                str[j] = k + 'a';                break;            }        }        cout << str << endl;    }    return 0;}

D：
/*
子集的组成只可能有两种
一种是1，和一个合法的非1数
一种是两个合法的非1数

可以用奇数偶数分类的办法证明不可能存在三个合法的非1数

#include <bits/stdc++.h>using namespace std;const int MAXN = 2e6 + 5;bool prime[MAXN];void init(){    for(int i = 0 ; i < MAXN ; i++) prime[i] = true;    prime[0] = prime[1] = false;    for(int i = 2 ; i < MAXN ; i++) {        if(prime[i] == false) continue;        for(int j = i + i ; j < MAXN ; j += i)             prime[j] = false;    }}int n;struct Node{    int u, id;}a[1000 + 5];bool cmp(Node a, Node b) {return a.u < b.u;}int main(){    init();    while(scanf("%d", &n) != EOF) {        for(int i = 1 ; i <= n ; i++) scanf("%d", &a[i].u), a[i].id = i;//        puts("zero");        sort(a + 1, a + 1 + n, cmp);//        puts("zero2");        int mark = n + 1;        int zeronum = 0;        for(int i = 1 ; i <= n ; i++) {            if(a[i].u != 1) {                mark = i;                break;            }            else zeronum++;        }        int ans1 = zeronum;        int re = -1;        for(int i = mark ; i <= n ; i++) if(prime[1 + a[i].u]) {ans1++; re = a[i].u; break;}//        puts("first");        int ans2 = 0;        int re2, re3;        re2 = re3 = -1;        for(int i = mark ; i <= n ; i++) {            for(int j = i + 1 ; j <= n ; j++) {                if(prime[a[i].u + a[j].u]) {                    ans2 = 2;                    re2 = a[i].u, re3 = a[j].u;                }            }        }//        puts("second");        int ans3 = 1;        if(ans1 >= ans2 && ans1 >= ans3) {            printf("%d\n", ans1);            int f = 1;            for(int i = 1 ; i < ans1 ; i++) printf("%d ", 1);            if(re != -1) printf("%d", re);            else printf("%d", 1);            puts("");        }        else if(ans2 >= ans1 && ans2 >= ans3) {            printf("%d\n", ans2);            printf("%d %d\n", re2, re3);        }        else {            printf("%d\n%d\n", ans3, a[1].u);        }    }    return 0;}

E：
/*
存一个之前所有数的异或和
由异或和的性质知道sum[l,r] = sum[1,l - 1] ^ sum[1,r]

然后用字典树存储这个异或和的二进制表示每次查询后插入即可

#include <bits/stdc++.h>using namespace std;#define LL long longconst int MAX = 31;const int MAXN = 1e6 + 5;const int MAXM = MAX * MAXN;int num[MAXM];int ne[MAXM][2];int cnt;int n, k;string toS(int u){    string str = "";    while(u) {        if(u & 1) str += '1';        else str += '0';        u /= 2;    }    while(str.size() < MAX - 1) str += '0';    reverse(str.begin(), str.end());    return str;}int query(string str, string st){    int now = 0;    int ans = 0;    int mark = 0;    while(now < str.size() && mark != -1) {        if(str[now] == '0' && st[now] == '0') {            if(ne[mark][1] != -1) ans += num[ne[mark][1]];            mark = ne[mark][0];        }        else if(str[now] == '0' && st[now] == '1') {            mark = ne[mark][1];        }        else if(str[now] == '1' && st[now] == '0') {            if(ne[mark][0] != -1) ans += num[ne[mark][0]];            mark = ne[mark][1];        }        else if(str[now] == '1' && st[now] == '1') {            mark = ne[mark][0];        }        now++;    }    if(mark != -1)        ans += num[mark];    return ans;}void ins(string str){    int u = 0;    for(int i = 0 ; i < str.size() ; i++) {        int j = str[i] - '0';        if(ne[u][j] == -1) ne[u][j] = ++cnt;        num[u]++;//        printf("%d ", u);        u = ne[u][j];    }    num[u]++;//    puts("");}int main(){    while(scanf("%d%d", &n, &k) != EOF) {        memset(num, 0, sizeof num);        memset(ne, -1, sizeof ne);        long long ans = 0;        cnt = 0;        int sum = 0;        string st = toS(k);        for(int i = 0 ; i <= n ; i++) {            int u;            if(i > 0)                scanf("%d", &u);            else u = 0;            u = u ^ sum;            string str = toS(u);            int temp = query(str, st);//            printf("i = %d, temp = %d, u = %d\n", i, temp, u);//            cout << str << endl;            ans = ans + temp;//            cout << str << endl;            ins(str);            sum = u;        }        printf("%I64d\n", ans);    }    return 0;}

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