东圃手机网站建设电话,天堂w区服选择,做网站怎么选关键词,新钥匙建站文章目录 Problem A. 逆序对染色#xff08;思维树状数组#xff09;Problem B. 连接召唤#xff08;贪心#xff09;Problem E. L 型覆盖检查器#xff08;模拟#xff09;Problem F. 小球进洞#xff1a;平面版#xff08;几何#xff09;Problem G. 函数查询Proble… 文章目录 Problem A. 逆序对染色思维树状数组Problem B. 连接召唤贪心Problem E. L 型覆盖检查器模拟Problem F. 小球进洞平面版几何Problem G. 函数查询Problem H. GG 和 YY 的石子游戏签到Problem L. 毛肚下清汤签到 Problem A. 逆序对染色思维树状数组
难似我了
#include bits/stdc.husing namespace std;#define int long longstruct BIT
{const int n;vectorint tree;BIT(int n) : n(n), tree(n 1) {};// 询问前x个数的和int Query(int x){int res 0;for (int i x; i 0; i - (i -i)) res tree[i];return res;}// 第l个位置zvoid Modify(int l, int z){if (l 0) return;for (int i l; i n; i (i -i)) tree[i] z;}// 区间求和int rangeQuery(int l, int r){return Query(min(n, r)) - Query(max(0ll, l - 1));}
};void solve()
{int n; cin n;vectorint a(n 1), pos(n 1);for (int i 1; i n; i ){cin a[i];pos[a[i]] i;}vectorvectorint mem(n 1);vectorbool st(n 1);for (int i 1; i n; i ){int t pos[a[i] - 1] 1;if (i t){mem[t].push_back(i);st[a[i]] true;}}BIT bit(n);int ans 0;for (int i 1; i n; i ){for (auto t : mem[i]) bit.Modify(a[t], 1);if (st[a[i]]) bit.Modify(a[i], -1);if (a[i - 1] 1 a[i] - 1) ans bit.rangeQuery(a[i - 1] 1, a[i] - 1);}cout ans \n;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t 1;// cin t;while (t -- ){solve();}
}Problem B. 连接召唤贪心
贪心优先1-5,2-4,3-3匹配我的代码里单独考虑了出现23的情况题解好像简单一些但是能a的代码就是好代码 我就不重写了
#include bits/stdc.husing namespace std;#define int long longtypedef pairint, int PII;void solve()
{vectorint cnt(6);for (int i 1; i 5; i ) cin cnt[i];int ans 0;int minn min(cnt[1], cnt[5]);ans minn;cnt[1] - minn; cnt[5] - minn;minn min(cnt[2], cnt[4]);ans minn;cnt[2] - minn; cnt[4] - minn;ans cnt[3] / 2;cnt[3] % 2;auto cal [](int id){int need 6 - id;int nw 0;for (int i 1; i id; i ){if (!cnt[i]) continue;int minn min((nw cnt[i]) / need, cnt[id]);cnt[i] - (minn * need - nw); cnt[id] - minn; for (int j 1; j i; j ) cnt[j] 0;ans minn;nw cnt[i];}if (nw ! 0 cnt[id]){int need 6 - nw;int c need / id;if (cnt[id] c){need - c * id;if (need 0){cnt[id] - c;ans ;for (int i 1; i id; i ) cnt[i] 0;}else{if (cnt[id] need){cnt[id] - need;ans ;for (int i 1; i id; i ) cnt[i] 0;}}}}if (id 5) ans cnt[id] / 2;else if (id 4 || id 2) ans cnt[id] / 3;else if (id 1) ans cnt[id] / 6;cnt[id] 0;};if (cnt[1] cnt[2] cnt[3]){cnt[1] - 1; cnt[2] - 1; cnt[3] - 1;ans ;if (cnt[2]) cal(2);if (cnt[1]) cal(1);}else if (cnt[2] cnt[3] cnt[5]){cnt[5] -- ; cnt[3] -- ;ans ;if (cnt[5]) cal(5);if (cnt[2]) cal(2);}else if (cnt[2] cnt[3]){if (cnt[2] 2){cnt[2] - 2; cnt[3] -- ;ans ;}if (cnt[2]) cal(2);}else{for (int i 5; i 1; i -- ){if (cnt[i]) cal(i);}}cout ans \n;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t 1;cin t;while (t -- ){solve();}
}Problem E. L 型覆盖检查器模拟
#include bits/stdc.h
using namespace std;
#define int long longstring g[510];
pairint, int dx[] {{-1, 0}, {0, 1}, {0, -1}, {-1, 0}};
pairint, int dy[] {{0,1}, {1,0}, {1,0}, {0,-1}};
char fx[] {D, L, R, D};
char fy[] {L, U, U, R};void solve()
{int n, m;cin n m;for (int i 1; i n; i) {cin g[i];g[i] g[i];}if (g[1][m] ! .) {cout No endl;return;}vectorpairint, int cc;for (int i 1; i n; i) {for (int j 1; j m; j) {if (g[i][j] C)cc.push_back({i, j});}}vectorvectorint st(n 1, vectorint(m 1, 0));for (int i 0; i cc.size(); i) {int a cc[i].x, b cc[i].y;st[a][b] i 1;for (int j 0; j 4; j) {int x1 a dx[j].x, y1 b dx[j].y;int x2 a dy[j].x, y2 b dy[j].y;if (x1 1 || x1 n || x2 1 || x2 n)continue;if (y1 1 || y1 m || y2 1 || y2 m)continue;if (st[x1][y1] ! 0) continue;if (st[x2][y2] ! 0) continue;if (g[x1][y1] fx[j] g[x2][y2] fy[j]) {st[x1][y1] i 1, st[x2][y2] i 1;}}}bool flag 0;mapint, int jk;for (int i 1; i n; i) {for (int j 1; j m; j) {if (i 1 j m) continue;if (st[i][j] 0){cout No endl;return;}jk[st[i][j]];}}for (auto bb : jk) {if (bb.y ! 3) {cout No endl;return;}}cout Yes endl;
}signed main()
{ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);int t 1;cin t;while (t--){solve();}
}
Problem F. 小球进洞平面版几何
待补
Problem G. 函数查询
待补
Problem H. GG 和 YY 的石子游戏签到
#includebits/stdc.h
using namespace std;
int t,m,ans;
long long n,a;
int main()
{scanf(%d,t);for(int i1;it;i){cinn;ans0;if(n%30){ans1;}a(n/3)n%3;coutans a\n;}return 0;
}Problem L. 毛肚下清汤签到
#include bits/stdc.husing namespace std;#define int long longtypedef pairint, int PII;void solve()
{int n; cin n;priority_queuePII, vectorPII, greaterPII red, green;for (int i 1; i n; i ){int a, b, c, d; cin a b c d;if (c 0 d 0) continue;else if (c 1 d 0) red.push({a, i});else if (c 0 d 1) green.push({b, i});else{if (a b) green.push({b, i});else red.push({a, i});}}cout red.size() ;while (red.size()){auto t red.top();red.pop();cout t.second ;}cout \n;cout green.size() ;while (green.size()){auto t green.top();green.pop();cout t.second ;}cout \n;
}signed main()
{ios::sync_with_stdio(false);cin.tie(0), cout.tie(0);int t 1;// cin t;while (t -- ){solve();}
}
