2015 Multi-University Training Contest 3

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A
定义一个合法序列：元素下标奇、偶相间。
题意：给定n个元素，有两个操作——修改第i个元素；查询区间[L, R]合法序列的最大和。
思路：线段树区间合并啦，不过返回结构体好使。

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1#define M(l, r) (l + r) >> 1using namespace std;typedef long long LL;const int MAXN = 1e5 + 1;const int MAXM = 1e6 + 1;const int INF = 1e9 + 10;struct Ans {    LL oo, oe, ee, eo;};struct Tree {    int l, r;    Ans ans;};Tree tree[MAXN<<2];void PushUp(int o) {    tree[o].ans.oo = max(tree[ll].ans.oe + tree[rr].ans.oo, tree[ll].ans.oo + tree[rr].ans.eo);    tree[o].ans.oo = max(tree[o].ans.oo, max(tree[ll].ans.oo, tree[rr].ans.oo));    tree[o].ans.oe = max(tree[ll].ans.oo + tree[rr].ans.ee, tree[ll].ans.oe + tree[rr].ans.oe);    tree[o].ans.oe = max(tree[o].ans.oe, max(tree[ll].ans.oe, tree[rr].ans.oe));    tree[o].ans.ee = max(tree[ll].ans.eo + tree[rr].ans.ee, tree[ll].ans.ee + tree[rr].ans.oe);    tree[o].ans.ee = max(tree[o].ans.ee, max(tree[ll].ans.ee, tree[rr].ans.ee));    tree[o].ans.eo = max(tree[ll].ans.eo + tree[rr].ans.eo, tree[ll].ans.ee + tree[rr].ans.oo);    tree[o].ans.eo = max(tree[o].ans.eo, max(tree[ll].ans.eo, tree[rr].ans.eo));}void Build(int o, int l, int r) {    tree[o].l = l; tree[o].r = r;    if(l == r) {        LL v; scanf("%lld", &v);        if(l & 1) {            tree[o].ans.oo = v;            tree[o].ans.oe = tree[o].ans.ee = tree[o].ans.eo = -1e15;        }        else {            tree[o].ans.ee = v;            tree[o].ans.oe = tree[o].ans.oo = tree[o].ans.eo = -1e15;        }        return ;    }    int mid = M(l, r);    Build(ll, l, mid); Build(rr, mid+1, r);    PushUp(o);}void Update(int o, int pos, LL v) {    if(tree[o].l == tree[o].r) {        if(tree[o].l & 1) tree[o].ans.oo = v;        else tree[o].ans.ee = v;        return ;    }    int mid = M(tree[o].l, tree[o].r);    if(pos <= mid) Update(ll, pos, v);    else Update(rr, pos, v);    PushUp(o);}Ans Best(Ans a, Ans b) {    Ans c;    c.oo = max(a.oe + b.oo, a.oo + b.eo);    c.oo = max(c.oo, max(a.oo, b.oo));    c.oe = max(a.oo + b.ee, a.oe + b.oe);    c.oe = max(c.oe, max(a.oe, b.oe));    c.ee = max(a.eo + b.ee, a.ee + b.oe);    c.ee = max(c.ee, max(a.ee, b.ee));    c.eo = max(a.eo + b.eo, a.ee + b.oo);    c.eo = max(c.eo, max(a.eo, b.eo));    return c;}Ans Query(int o, int L, int R) {    if(tree[o].l == L && tree[o].r == R) {        return tree[o].ans;    }    int mid = M(tree[o].l, tree[o].r);    if(R <= mid) {        return Query(ll, L, R);    }    else if(L > mid) {        return Query(rr, L, R);    }    else {        return Best(Query(ll, L, mid), Query(rr, mid + 1, R));    }}int main(){    int t; scanf("%d", &t);    while(t--) {        int n, m; scanf("%d%d", &n, &m);        Build(1, 1, n);        while(m--) {            int op, x, y; scanf("%d%d%d", &op, &x, &y);            if(op == 0) {                Ans T = Query(1, x, y);                printf("%lld\n", max(max(max(T.oo, T.oe), T.ee), T.eo));            }            else {                Update(1, x, y);            }        }    }    return 0;}

B
定义：f[i]是i的质因子个数。
题意：给定1-n共n个元素和m次查询，每次查询区间[L, R]里面最大的gcd(f[i]，f[j)其中L<=i,j<=R。
思路：发现n的限制下，f[]值最大为7。那样我们统计一下前i个数里面f值为j出现多少次即可。

这样时间复杂度是O(7∗7∗T)

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int MAXN = 1e6 + 1;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;int gcd(int a, int b) {    return b == 0 ? a : gcd(b, a % b);}bool vis[MAXN];int g[8][8];int a[MAXN];int sum[MAXN][8];int main(){    for(int i = 2; i < MAXN; i++) {        if(vis[i]) continue;        a[i] = 1;        for(int j = 2 * i; j < MAXN; j += i) {            vis[j] = true;            a[j]++;        }    }    for(int i = 1; i < MAXN; i++) {        for(int j = 1; j <= 7; j++) {            if(j == a[i]) {                sum[i][j] = sum[i-1][j] + 1;            }            else {                sum[i][j] = sum[i-1][j];            }        }    }    for(int i = 1; i <= 7; i++) {        for(int j = i + 1; j <= 7; j++) {            g[i][j] = gcd(i, j);        }    }    int t; scanf("%d", &t);    while(t--) {        int L, R; scanf("%d%d", &L, &R);        int ans = 1;        for(int i = 1; i <= 7; i++) {            vis[i] = false;            int num = sum[R][i] - sum[L-1][i];            if(num > 1) {                ans = max(ans, i);            }            if(num) vis[i] = true;        }        for(int i = 1; i <= 7; i++) {            if(!vis[i]) continue;            for(int j = i + 1; j <= 7; j++) {                if(!vis[j]) continue;                ans = max(ans, g[i][j]);            }        }        printf("%d\n", ans);    }    return 0;}

D
题意：R只能沿着主对角线涂，B只能沿着副对角线涂，R、B都涂的地方为G，其中每个R、B只被涂了一次，.表示未被涂。问最少需要涂多少次。

模拟即可。

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int MAXN = 1e6 + 1;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;char str[60][60];bool vis[60][60];int n, m, ans;int go[2][2] = {1, -1, 1, 1};bool judge(int x, int y) {    return x >= 0 && x < n && y >= 0 && y < m;}void DFS(int x, int y, int d, char op) {    str[x][y] = '.';    int nx = x + go[d][0];    int ny = y + go[d][1];    if(judge(nx, ny) && str[nx][ny] == op) {        DFS(nx, ny, d, op);    }    else {        ans++;    }}int main(){    int t; scanf("%d", &t);    while(t--) {        scanf("%d", &n);        for(int i = 0; i < n; i++) {            scanf("%s", str[i]);            m = strlen(str[i]);        }        CLR(vis, false);        for(int i = 0; i < n; i++) {            for(int j = 0; j < m; j++) {                if(str[i][j] == 'G') {                    str[i][j] = 'R';                    vis[i][j] = true;                }            }        }        ans = 0;        for(int i = 0; i < n; i++) {            for(int j = 0; j < m; j++) {                if(str[i][j] == 'R') {                    DFS(i, j, 1, 'R');                    //cout << i << ' ' << j << ' ' << ans << endl;                }            }        }        for(int i = 0; i < n; i++) {            for(int j = 0; j < m; j++) {                if(vis[i][j]) {                    str[i][j] = 'B';                }            }        }        for(int i = 0; i < n; i++) {            for(int j = 0; j < m; j++) {                if(str[i][j] == 'B') {                    DFS(i, j, 0, 'B');                    //cout << i << ' ' << j << ' ' << ans << endl;                }            }        }        printf("%d\n", ans);    }    return 0;}

题意：0-n线段树，给定一个节点所代表的区间[L, R]问你包含该节点的线段树是否存在，存在输出最小的n值。

DFS一下即可，显然到某个程度就break了，不确定就只好试试了。。。

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int MAXN = 1e6 + 1;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;LL ans = -1;void DFS(LL L, LL R, LL pl, LL pr, int T) {    //cout << L << ' ' << R << endl;    if(L < 0 || L > R) return ;    if(pl == L && pr == R) return ;    if(ans != -1 && R > ans) return ;    if(L == 0) {        if(ans == -1) ans = R;        else {            ans = min(ans, R);        }        return ;    }    if(T > 10) {        return ;    }    DFS(L, 2 * R - L, L, R, T + 1); DFS(L, 2 * R + 1 - L, L, R, T + 1);    DFS((L - 1) * 2 - R, R, L, R, T + 1); DFS((L - 1) * 2 + 1 - R, R, L, R, T + 1);}int main(){    LL L, R;    while(scanf("%lld%lld", &L, &R) != EOF) {        if(L == R || L == 0) {            printf("%lld\n", R);            continue;        }        ans = -1; DFS(L, R, -1, -1, 0);        printf("%lld\n", ans);    }    return 0;}

K
签到题

#include <cstdio>#include <algorithm>#include <iostream>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;const int MAXN = 1e6 + 1;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;vector<int> G[110];int son[110], d[110];int ans, k;void DFS(int u) {    son[u] = 1;    for(int i = 0; i < G[u].size(); i++) {        int v = G[u][i];        DFS(v);        son[u] += son[v];    }    ans += son[u] - 1 == k;}int main(){    int n;    while(scanf("%d%d", &n, &k) != EOF) {        for(int i = 1; i <= n; i++) {            G[i].clear(); d[i] = 0;        }        for(int i = 0; i < n - 1; i++) {            int u, v; scanf("%d%d", &u, &v);            G[u].push_back(v); d[v]++;        }        int s;        for(int i = 1; i <= n; i++) {            if(d[i] == 0) {                s = i; break;            }        }        ans = 0; DFS(s);        printf("%d\n", ans);    }    return 0;}

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