ACM-ICPC国际大学生程序设计竞赛北京赛区(2017)网络赛

A

连续的k个要有准确的m个可用的天数，并且第一个可以用的 + 另外一个要求值最小

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;const int INF = 1e9 + 10;int num[105];bool istra[105];int n, m, p;struct Node {int tmp, id;bool operator < (const Node &w) const {return tmp < w.tmp;}}node[105];int main(){while(scanf("%d%d", &n, &m) != EOF) {for(int i = 0; i < n; ++i) {scanf("%d", num + i);}scanf("%d", &p);mst(istra, false);for(int x, i = 0; i < p; ++i) {scanf("%d", &x);istra[x] = true;}int minx = INF, a = 0, b = 1;int day = 0;for(int len = m; len <= n; ++len) {for(int i = 0; i < n; ++i) {int l = i, r = i + len - 1;if(r >= n)continue;int cnt = 0;int xday = 0;for(int j = l; j <= r; ++j) {if(istra[j]) continue;node[cnt].tmp = num[j];node[cnt].id = j;xday++, cnt++;}if(xday != m)continue;int x = node[0].tmp, idx = node[0].id;sort(node + 1, node + cnt);int y = node[1].tmp;int idy = node[1].id;if(x + y < minx) {minx = x + y;a = idx, b = idy;}}}if(a > b)swap(a, b);printf("%d %d\n", a, b);}return 0;}

C

参考了大神的思路：blackcat

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 300 + 10;const int INF = 1e9 + 10;int dp[qq][2], a[qq][qq];int sum[qq], val[qq];int Solve(int m, int p) {int ans = -INF;for(int i = 1; i <= m; ++i) {int minx = INF;int sumX = 0;for(int j = i; j <= m; ++j) {minx = min(minx, val[j]);sumX += sum[j];if(i == 1 && j == m) {ans = max(ans, sumX - minx + p);} else {int tsumX = max(sumX - minx + p, sumX);ans = max(ans, tsumX);}}}return ans;}int Dp(int m, int p) {dp[1][0] = sum[1];dp[1][1] = sum[1] - val[1] + p;for(int i = 1; i <= m; ++i) {dp[i][0] = max(dp[i - 1][0], 0) + sum[i];dp[i][1] = max(dp[i - 1][1] + sum[i], max(dp[i - 1][0], 0) - val[i] + p + sum[i]);}int ans = -INF;for(int i = 1; i <= m; ++i) {ans = max(ans, max(dp[i][0], dp[i][1]));}return ans;}int main(){int n, m, p;while(scanf("%d%d%d", &n, &m, &p) != EOF) {for(int i = 1; i <= n; ++i) {for(int j = 1; j <= m; ++j) {scanf("%d", &a[i][j]);}}int ans = -INF;for(int i = 1; i <= n; ++i) {fill(sum + 1, sum + 1 + m, 0);fill(val + 1, val + 1 + m, INF);for(int j = i; j <= n; ++j) {for(int k = 1; k <= m; ++k) {sum[k] += a[j][k];val[k] = min(val[k], a[j][k]);}if(i == 1 && j == n) {ans = max(ans, Solve(m, p));} else {ans = max(ans, Dp(m, p));}}}printf("%d\n", ans);}return 0;}

E

首先排除n <= 3的情况， n = 3的时候特判一下是否三点共线，不共线就是NO，n > 3的情况的话肯定是YES，我们可以先判断是否存在三点共线，如果存在就把中间一个点的染成B其他点全部染成A即可，如果没有三点贡献的情况，就用跑一下凸包把外围的点全部标记一下，然后找出中间一个点染成B其他点染成A即可，如果中间没有点的话，就这样任意找两点连成一条直线，判断这条直线左边和右边是否都存在点，如果都存在就把这个点染成B，其他点染成A

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int qq = 1e5 + 10;const int INF = 1e9 + 10;int n;struct Point {int x, y;int id;Point(){}Point(int _x, int _y) {x = _x, y = _y;}bool operator < (const Point &w) const {if(x == w.x)return y < w.y;return x < w.x;}Point operator - (const Point &w) const {return Point(x - w.x, y - w.y);}int operator ^ (const Point &w) const {return x * w.y - y * w.x;}int operator * (const Point &w) const {return x * w.x + y * w.y;}}p[105], tmp[103];bool vis[105];char ans[105];int Xmult(Point p0, Point p1, Point p2) {return (p1 - p0) ^ (p2 - p0);}int main(){intt;scanf("%d", &t);while(t--) {mst(ans, 0);scanf("%d", &n);for(int i = 1; i <= n; ++i) {scanf("%d%d", &p[i].x, &p[i].y);p[i].id = i;}if(n < 3) {puts("NO");continue;}if(n == 3) {sort(p + 1, p + 1 + 3);int a = p[3].y - p[1].y, b = p[3].x - p[1].x;int c = p[2].y - p[1].y, d = p[2].x - p[1].x;if(a * d == b * c) {puts("YES");ans[p[1].id] = ans[p[3].id] = 'A';ans[p[2].id] = 'B';for(int i = 1; i <= 3; ++i) {printf("%c", ans[i]);} puts("");} else {puts("NO");}continue;}puts("YES");bool f = false;for(int i = 1; i <= n; ++i) {for(int j = i + 1; j <= n; ++j) {int a = p[j].y - p[i].y, b = p[j].x - p[i].x;for(int k = j + 1; k <= n; ++k) {int c = p[k].y - p[i].y, d = p[k].x - p[i].x;if(a * d == b * c) {f = true;tmp[0] = p[i], tmp[1] = p[j], tmp[2] = p[k];}}if(f)break;}if(f)break;}if(f) {sort(tmp, tmp + 3);ans[tmp[1].id] = 'B';for(int i = 1; i <= n; ++i) {if(ans[i] != 'B')ans[i] = 'A';printf("%c", ans[i]);}puts("");} else {mst(vis, false);for(int i = 1; i <= n; ++i) {for(int j = 1; j <= n; ++j) {int a, b;a = b = 0;for(int k = 1; k <= n; ++k) {if(k == i || k == j)continue;if(Xmult(p[k], p[i], p[j]) > 0)a++;elseb++;}if(a == n - 2)vis[i] = vis[j] = true;}}bool flag = false;for(int i = 1; i <= n; ++i) {if(!vis[i]) {ans[i] = 'B';flag = true;break;}}if(flag) {for(int i = 1; i <= n; ++i) {if(ans[i] != 'B')ans[i] = 'A';printf("%c", ans[i]);}puts("");continue;}for(int i = 1; i <= n; ++i) {for(int j = i + 1; j <= n; ++j) {int a, b;a = 0, b = 0;for(int k = 1; k <= n; ++k) {if(k == i || k == j)continue;if(Xmult(p[k], p[i], p[j]) > 0)a++;elseb++;}if(a > 0 && b > 0) {flag = true;ans[i] = 'B', ans[j] = 'B';}if(flag)break;}if(flag)break;}for(int i = 1; i <= n; ++i) {if(ans[i] != 'B')ans[i] = 'A';printf("%c", ans[i]);}puts("");}}return 0;}

G

队友写的

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;LL gcd(LL x,LL y) {    return y==0?x:gcd(y,x%y);}int main() {    LL n,m;    while(~scanf("%lld%lld",&n,&m)) {        LL x=n-1;        LL y=m-1;        LL g=gcd(x,y);        LL a=x/g;        LL b=y/g;        LL ret=x*y/g;        ret-=(ret/g)-a-b;        printf("%lld\n",ret);    }    return 0;}

I

因为x和y可以相等，无非就找一下三种情况的最小值，第一种区间全正一定是最小正数的平方最小，第二种区间全负一定是最大负数的平方最小，第三种区间有正有负，一定是最大正数和最小负数的乘积最小，线段树维护一下四种值即可

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <ctime>#include <iostream>#include <algorithm>#include <sstream>#include <string>#include <vector>#include <queue>#include <stack>#include <map>#include <set>#include <utility>#include <bitset>using namespace std;#define LL long long#define pb push_back#define mk make_pair#define pill pair<int, int>#define mst(a, b)memset(a, b, sizeof a)#define REP(i, x, n)for(int i = x; i <= n; ++i)const int MOD = 1e9 + 7;const int INF = 1e9 + 10;const LL MAXN = 1e17;const int qq = 1 << 17;int pminx[qq << 2], pmaxn[qq << 2];int nminx[qq << 2], nmaxn[qq << 2];void Init(int rt) {pminx[rt] = INF, pmaxn[rt] = 0;nminx[rt] = 0, nmaxn[rt] = -INF;}void PushUp(int rt) {pminx[rt] = min(pminx[rt << 1], pminx[rt << 1 | 1]);pmaxn[rt] = max(pmaxn[rt << 1], pmaxn[rt << 1 | 1]);nminx[rt] = min(nminx[rt << 1], nminx[rt << 1 | 1]);nmaxn[rt] = max(nmaxn[rt << 1], nmaxn[rt << 1 | 1]);}void Build(int rt, int l, int r) {Init(rt);if(l == r) {int x;scanf("%d", &x);if(x > 0)pminx[rt] = pmaxn[rt] = x;elsenminx[rt] = nmaxn[rt] = x;return;}int m = (l + r) >> 1;Build(rt << 1, l, m);Build(rt << 1 | 1, m + 1, r);PushUp(rt);}void UpDate(int rt, int l, int r, int x, int y, int val) {if(l == r) {Init(rt);if(val > 0)pminx[rt] = pmaxn[rt] = val;elsenminx[rt] = nmaxn[rt] = val;//printf("%d %d %d %d\n", pmaxn[rt], pminx[rt], nmaxn[rt], nminx[rt]);return;}int m = (l + r) >> 1;if(x <= m)UpDate(rt << 1, l, m, x, y, val);if(y > m)UpDate(rt << 1 | 1, m + 1, r, x, y, val);PushUp(rt);}LL Query(int rt, int l, int r, int x, int y, int f) {LL ans;if(f == 1)ans = 0;else if(f == 2)ans = INF;else if(f == 3)ans = -INF;elseans = 0;if(x <= l && r <= y) {if(f == 1)return (LL)pmaxn[rt];else if(f == 2)return (LL)pminx[rt];else if(f == 3)return (LL)nmaxn[rt];elsereturn (LL)nminx[rt];}int m = (l + r) >> 1;if(x <= m) {if(f == 1)ans = max(ans, Query(rt << 1, l, m, x, y, f));else if(f == 2)ans = min(ans, Query(rt << 1, l, m, x, y, f));else if(f == 3)ans = max(ans, Query(rt << 1, l, m, x, y, f));elseans = min(ans, Query(rt << 1, l, m, x, y, f));}if(y > m) {if(f == 1)ans = max(ans, Query(rt << 1 | 1, m + 1, r, x, y, f));else if(f == 2)ans = min(ans, Query(rt << 1 | 1, m + 1, r, x, y, f));else if(f == 3)ans = max(ans, Query(rt << 1 | 1, m + 1, r, x, y, f));elseans = min(ans, Query(rt << 1 | 1, m + 1, r, x, y, f));}return ans;}int main(){int t;scanf("%d", &t);while(t--) {int n;scanf("%d", &n);n = 1 << n;Build(1, 1, n);int Q;scanf("%d", &Q);while(Q--) {int op, l, r;scanf("%d%d%d", &op, &l, &r);l++, r++;if(op == 2) {UpDate(1, 1, n, l, l, r - 1);} else {LL maxnz = Query(1, 1, n, l, r, 1);LL minxz = Query(1, 1, n, l, r, 2);LL maxnf = Query(1, 1, n, l, r, 3);LL minxf = Query(1, 1, n, l, r, 4);//printf("%lld %lld %lld %lld\n", maxnz, minxz, maxnf, minxf);LL ans = MAXN;if(maxnf != -INF) {ans = min(ans, maxnf * maxnf);}if(minxz != INF) { ans = min(ans, minxz * minxz);}if(minxf != 0 && maxnz != 0) {ans = min(ans, minxf * maxnz);}printf("%lld\n", ans);}}}return 0;}