Codeforces contest 787 recordings

比赛记录

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A - The Monster

Problem Description

求i，使得x=i×a+b且x=j×c+d

Data Limit

a,b,c,d<100

Solution

强行模拟、exgcd均可做

Code

#include<cstdio>#include<iostream>using namespace std;int main() {    int a, b, c, d;    scanf("%d%d%d%d", &a, &b, &c, &d);    for(int i = 0; i <= 10000; i ++) {        while(b < d) b += a;        while(b > d) d += c;        if(b == d) {            printf("%d\n", b);            return 0;        }    }    printf("-1\n");    return 0;}

B - Not Afraid

Problem Description

给m个数列，若每个数列中均同时有一个数及其相反数，则输出YES

Data Limit

∑mi=1ki≤104

Solution

模拟

Code

#include <cstdio>#include <iostream>#include <cstring>using namespace std;bool tab[50005];int main() {    int n, m, k, x;    scanf("%d%d", &n, &m);    for (int i = 1; i <= m; i ++) {        memset(tab, 0, sizeof(tab));        scanf("%d", &k);        bool flag = 0;        while (k --) {            scanf("%d", &x);            if (tab[20000 - x]) flag = 1;            tab[20000 + x] = 1;        }        if (!flag) {            printf("YES\n");            return 0;        }    }    printf("NO\n");    return 0;}

赛后补题

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C - Berzerk

Problem Description

两个人博弈，棋盘是一个n个点的环，每人有序列s，可以使棋子顺时针移动s[i][j]个单位，使棋子移
动到第一格的人获胜，求对于棋子的每一个起始位置，两个人分别作为先手，是否必胜

Data Limit

n<=7000

Solution

考虑每个状态f[i][j]表示走到第i个位置，先手是j时先手的是否必胜，f[i][j]可以转移到
f[(i+s[k][j])][! j]，若它可以转移到的每一个状态都是对手必胜，则该状态为必败，若
该状态可以转移到任何一个对手必败的局面，则该局面必胜，否则可以导致循环

Code

#include <cstdio>#include <cstring>#include <iostream>using namespace std;const int N = 7005;int s[N][2], f[N][2], n, k[2];int vis[N][2];inline int work(int pos, int fir) {    if (vis[pos][fir] || f[pos][fir]) return f[pos][fir];    vis[pos][fir] = 1;    int nxt, val;    for (int i = 1; i <= k[fir]; i ++) {        nxt = (pos + s[i][fir]) % n;        if (!nxt) nxt = n;        if (nxt == 1) {            f[pos][fir] = 1;            return 1;        }    }    int cnt = 0;    for (int i = 1; i <= k[fir]; i ++) {        nxt = (pos + s[i][fir]) % n;        val = work(nxt, 1 - fir);        if (val == -1) {            f[pos][fir] = 1;            return 1;        }        else if(val == 1) cnt ++;    }    if (cnt == k[fir]) f[pos][fir] = -1;    return f[pos][fir];}int main() {    scanf("%d", &n);    for (int j = 0; j <= 1;j ++) {        scanf("%d", &k[j]);        for (int i = 1; i <= k[j]; i ++) scanf("%d", &s[i][j]);    }    for (int j = 0; j <= 1; j ++) {        memset(vis, 0, sizeof(vis));        for (int i = n; i >= 1; i --) {            work(i, 0);            work(i, 1);        }    }    for(int j = 0; j <= 1; j ++) {        for(int i = 2; i <= n; i ++) {            if (f[i][j] == -1) printf("Lose ");            else if (f[i][j] == 0) printf("Loop ");            else printf("Win ");        }        printf("\n");    }    return 0;}

D - Legacy

Problem Description

单源最短路，允许u−>v或u−>[l,r]或[l,r]−>u

Data Limit

1 ≤ n, q ≤ 105, 1 ≤ s ≤ n

Solution

我们试图用一些点代表一些区间，于是我们得到了线段树

于是我们顺利地把边数变成了O(log(q))的级别

Code

#include <cstdio>#include <iostream>#include <queue>#include <cstring>using namespace std;typedef long long LL;const int N = 100005;struct node{int nxt, to, w;} edge[3100000];int que[N], cnt = 0, tot;int fir[N * 8], back[N * 8];LL dis[N * 8];bool inq[N * 8];inline void add(int x, int y, int l) {    edge[++ cnt].to = y;    edge[cnt].nxt = fir[x];    edge[cnt].w = l;    fir[x] = cnt;}inline void build(int x, int l, int r) {    if (l == r) {        back[l] = x;        add(x, x + N * 4, 0);        add(x + N * 4, x, 0);        return;    }    int mid = (l + r) >> 1, lc = x << 1, rc = lc + 1;    build(lc, l, mid);    build(rc, mid + 1, r);    add(x, lc, 0);    add(x, rc, 0);    add(lc + N * 4, x + N * 4, 0);    add(rc + N * 4, x + N * 4, 0);}inline void Query(int x, int l, int r, int a, int b) {    if (l == a && r == b) {        que[++ tot] = x;        return;    }    int mid = (l + r) >> 1;    if (b <= mid) Query(x << 1, l, mid, a, b);    else if (a > mid) Query(x << 1 | 1, mid + 1, r, a, b);    else Query(x << 1, l, mid , a, mid), Query(x << 1 | 1, mid + 1, r, mid + 1, b);}int main() {    int n, q, s, opt, v, l, r, u, w;    scanf("%d%d%d", &n, &q, &s);    build(1, 1, n);    while (q --) {        scanf("%d", &opt);        if (opt == 1) {            scanf("%d%d%d", &u, &v, &w);            add(back[u], back[v], w);        } else {            scanf("%d%d%d%d", &v, &l, &r, &w);            tot = 0;            Query(1, 1, n, l, r);            if (opt == 2) {                for (int i = 1; i <= tot; i ++) add(back[v], que[i], w);            } else {                for (int i = 1; i <= tot; i ++) add(que[i] + N * 4, back[v], w);            }        }    }    memset(dis, 127, sizeof(dis));    LL un = dis[0];    queue<int> Q;    Q.push(back[s]);    dis[back[s]] = 0;    while (!Q.empty()) {        int x = Q.front();        Q.pop();        inq[x] = 0;        for (int i = fir[x]; i; i = edge[i].nxt) {            v = edge[i].to;            if(dis[v] > dis[x] + (LL)edge[i].w) {                dis[v] = dis[x] + (LL)edge[i].w;                if (!inq[v]) {                    inq[v] = 1;                    Q.push(v);                }            }        }    }    for (int i = 1; i <= n; i ++) {        if(dis[back[i]] != un) printf("%I64d ", dis[back[i]]);        else printf("-1 ");    }    printf("\n");    return 0;}

E - Till I Collapse

Problem Description

对于一个序列，求k∈[1,n]，至少将序列分成几块，使每块内不同的数总数不超过k

Data Limit

n≤105

Solution

经过观察发现答案数组是非严格递减的，那么可以知道如果对于一个区间[l,r]中，如果
ans[l]==ans[r]那么整个区间ans应该是一样的。所以把这种情况剪枝一下，然后就过了
标算是可持久化数据结构

Code

#include <cstdio>#include <iostream>#include <cstring>using namespace std;const int N = 100005;int ans[N], col[N], tab[N], n;inline int count(int x) {    memset(tab, 0, sizeof(tab));    int ret = 1, cnt = 0;    for (int i = 1; i <= n; i ++) {        if (tab[col[i]] == ret) continue;        tab[col[i]] = ret;        cnt ++;        if (cnt > x) {            cnt = 1;            tab[col[i]] = ++ ret;        }    }    return ret;}inline void work(int l, int r) {    if (l > r) return;    int cl = count(l);    if (l == r) {        ans[l] = cl;        return;    }    int cr = count(r);    if (cl == cr) {        for (int i = l; i <= r; i ++) ans[i] = cl;        return;    }    ans[l] = cl;    ans[r] = cr;    int mid = (l + r) >> 1;    work(l + 1, mid);    work(mid + 1, r - 1);}int main() {    scanf("%d", &n);    for (int i = 1; i <= n; i ++) scanf("%d", &col[i]);    work(1, n);    for (int i = 1; i <= n; i ++) printf("%d ",ans[i]);    printf("\n");    return 0;}