20201104大模拟（一）

开错题了我吐了，赛后A一发过，绝了

B是什么deng西，我得有一段时间不想看到植物大战僵尸了，有一些边界问题

C是在计算的过程中的大模拟

D是思维，

总体来说其实还好，不要怕的，就还好

“一道题WA了五发以上其实就没有出的必要了”

自己想想为什么会打成这个样子好不好

A - Try and Catch

题目真的有点长，还是要耐心阅读理解吧，

只有一个throw，刚开始以为有好多，并且觉得空格挺难搞的就先没做，其实没啥坑的直接冲就完事

#include <bits/stdc++.h>
using namespace std;struct node {int pos1;int pos2;string a;string b;
};stack<int> tryy;
vector<node> catchh; string throww = "";
int idx;bool cmp(node x, node y) {return x.pos1 < y.pos1;
}int main() {int n;cin >> n;getchar();for(int i = 1; i <= n; ++ i) {string s;getline(cin, s);int len = s.length();for(int j = 0; j < len; ++ j) {if(s[j] == ' ') continue;if(s[j] == 't') {if(s[j + 1] == 'r') { // trytryy.push(i);}else { // throwint k = j + 1;for(; k < len; ++ k) {if(s[k] == '(') break;}++ k;for(; k < len; ++ k) {if(s[k] != ' ') break;}for(; k < len; ++ k) {if(s[k] == ' ' || s[k] == ')') break;throww = throww + s[k];}idx = i;}break;}if(s[j] == 'c') { // catchstring aa = "";int k = j + 1;for(; k < len; ++ k) {if(s[k] == '(') break;}++ k;for(; k < len; ++ k) {if(s[k] != ' ') break;}for(; k < len; ++ k) {if(s[k] == ' ' || s[k] == ',') break;aa = aa + s[k];}string bb = "";for(; k < len; ++ k) {if(s[k] == '\"') break;}++ k;for(; k < len; ++ k) {if(s[k] == '\"') break;bb = bb + s[k];}//cout << i << " " << aa << " " << bb << endl;int pos1 = tryy.top();tryy.pop();catchh.push_back({pos1, i, aa, bb});break;}}}sort(catchh.begin(), catchh.end(), cmp);int siz = catchh.size();//    cout << "---- catchh :\n" ;
//  for(auto x : catchh) {//      cout << x.pos1 << " " << x.pos2 << " " << x.a << " " << x.b << endl;
//  }
//
//  cout << "\n\n";
//  cout << "idx :" << idx << endl;bool f = 0;for(int i = siz - 1; i >= 0; -- i) {if(f) break;node tmp = catchh[i];//cout << tmp.pos1 << " " << tmp.pos2 << endl;if(tmp.pos1 < idx && tmp.pos2 > idx) {//cout << throww << " " << tmp.a << endl; //cout << (tmp.a == throww) << endl;if(tmp.a == throww) {cout << tmp.b ;f = 1;break;}}}if(!f) {cout << "Unhandled Exception";}return 0;
}

B - I, zombie

植物大战僵尸，开错题目了，

一些边界问题

第一个是土豆雷（这个赛时就想到了（比较后面才想到的
有五十多个豌豆射手，或者说对于僵尸的血量可以是负的这件事的理解上面，是知道的，但是在处理过程中有个地方漏掉了，（

刚开始的思路是模拟每一段，每一段要么是MPPPPPPP 要么是MMMMM 就大概是两种，还有一些特殊情况比如PPPPP，但是其实可以归进去

对于每一段，他的左边有多少个豌豆，当前段要走几步什么的进行模拟（二分

但是关于僵尸血量可以是负的这件事，感觉可以换一种思路进行模拟，

还是要二分

最后一个位置是直接到达的，其他所有位置都要先走一步才能到达

MPMPMPMPMPMPMPM
12

60个豌豆还是处理错了

PPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPMMMMMMMMMM
82

还是上面这个60个豌豆的样例，如果是前面全是P, 尸体可以骗土豆雷，但是去吃脑子的时候不能是尸体~~（（？？有点不确定了，~~（对的，但是不是这里的问题

这是第二次重码了，准备开始第三次

WA 也许哪天有心情可以改一改

#include <bits/stdc++.h>
using namespace std;const int maxn = 110;
int sum[maxn];bool judge(int leff, int p, int x, bool f) {if(x == 0) return 0;//if(p == 0) return 1;if(!f) {int z = x;int hp = 50;if(p) {for(int i = p; i > 0; -- i) {hp -= leff + i;if(hp <= 0) {z --;hp = 50;}if(z <= 0) return 0;int pea = 10;while(pea > 0) {pea -= z;hp -= leff + i;if(hp <= 0){z --;hp = 50;}if(z <= 0) return 0;}}  }else {if(x >= 2) return 1;else return 0;}}else {int z = x;int hp = 50;if(p) {int pea = 10;while(pea > 0) {pea -= z;hp -= leff + p;if(hp <= 0){z --;hp = 50;}if(z <= 0) return 0;}    }else {if(x) return 1;else return 0;}for(int i = p - 1; i > 0; -- i) {hp -= leff + i;if(hp <= 0) {z --;hp = 50;}if(z <= 0) return 0;int pea = 10;while(pea > 0) {pea -= z;hp -= leff + i;if(hp <= 0){z --;hp = 50;}if(z <= 0) return 0;}} }return 1;
}int main () {int T;cin >> T;int cas = 0;while(T --) {++ cas;printf("Case %d: ", cas);string s;cin >> s;int n = s.size();if(s[0] == 'P') sum[0] = 1;else sum[0] = 0;for(int i = 1; i < n; ++ i) {sum[i] = sum[i - 1] + (s[i] == 'P' ? 1 : 0);}bool f = 1; int cntp = 0;int tot = 0;for(int i = n - 1; i >= 0; -- i) {if(s[i] == 'M') {cout << "--------- cntp :" << cntp << ' ' << f << endl;int l = 0, r = 200000, ans;while(l <= r) {int mid = (l + r) >> 1;if(judge(sum[i], cntp, mid, f)) {ans = mid;r = mid - 1;}else {l = mid + 1;}}cout << ans << endl; tot += ans;f = 0;cntp = 0;}else {cntp ++;}}//cout << tot << endl;//cout << cntp << endl;if(cntp > 0) {int l = 0, r = 200000, ans;while(l <= r) {int mid = (l + r) >> 1;if(judge(0, cntp, mid, f)) {ans = mid;r = mid - 1;}else {l = mid + 1; }}tot += ans;}if(s[0] == 'M') tot ++;cout << tot << endl;}
}

我要按网上思路来le，，可恶十一点了

0:17 按网上别人的代码流程来的，没啥意思说实话，不过这样写不容易出错，大体思路是一样的，

#include <bits/stdc++.h>
using namespace std;int step;bool judge(int pnum, int cnt, int znum) {int st = step;int php = 10, zhp = 50;while(cnt > 0 && znum > 0) {if(st > 0) {st --;zhp -= pnum;}else {php -= znum;zhp -= pnum;}if(php <= 0) {php = 10;pnum --; cnt --;st = 1;}if(zhp <= 0) {zhp = 50;znum --;}}if(cnt <= 0 && znum > 0) return 1;else return 0;
}int getRes(int pnum, int cnt) {int l = 1, r = 200000, ans;while(l <= r) {int mid = (l + r) >> 1;if(judge(pnum, cnt, mid)) {ans = mid;r = mid - 1;}else {l = mid + 1;}}return ans;
}int main() {int T;cin >> T;int cas = 0;while(T --) {++ cas;printf("Case %d: ", cas);string s;cin >> s;int n = s.size();int pnum = 0;int mnum = 0;for(int i = 0; i < n; ++ i) {if(s[i] == 'P') pnum ++;else mnum ++;}int ans = 0; -- n;if(s[n] == 'M') {ans ++;-- n;step = 2;}else step = 1;while(n >= 0) {if(s[n] == 'M') {if(step > 1) {step --;if(pnum >= 50) ans ++;continue;}-- n;mnum --;ans ++;step = 2;} else {int cnt =  0;for(int i = n; n >= 0; -- i) {if(s[i] == 'P') cnt ++;else break;}int res = getRes(pnum, cnt);//cout << pnum << " " << cnt << " " << res << endl;pnum -= cnt;ans += res;n -= cnt + 1;step = 2; }}if(s[0] == 'M') ans ++;cout << ans << endl;} return 0;
}

C - Broken Watch

刚开始没有题面（），其实不是很难的，

小数据手模找一下规律分奇偶推出两个公式

有个对2642 ^ {64}264的取模，unsigned long long 的自然溢出

然后有个除法，刚开始直接除然后wa了，后来考虑到有取模问题就需要用到逆元啥的，码了一个扩展欧几里得，但是发现2642^{64}264其实取不到啊,ull是264−12 ^ {64} - 1264−1

于是只能在做的过程中先除掉，分类分死我了

#include <bits/stdc++.h>
using namespace std;typedef unsigned long long ull;//ull mod;
//
//ull exgcd(ull a, ull b, ull & x, ull & y) {//  if(b == 0) {//      x = 1, y = 0;
//      return a;
//  }
//
//  ull d = exgcd(b, a % b, y, x);
//
//  y -= (a / b) * x;
//
//  return d;
//}int main() {ull a, b, c, n;cin >> a >> b >> c >> n;int cnt = 1;if(a == b && a == c) {cnt = 3;}else if(a == b || a == c || b == c){cnt = 2;}ull ans;if(n <= 2) {cout << 0 ;return 0;}if(n & 1) {if(cnt == 1)ans = n * (1ull + (n / 2)) * (n / 2) ;else if(cnt == 2) {if(!((n / 2) & 1)) {ans = n * (1ull + (n / 2)) * (n / 4);}else {ans = n * ((1ull + (n / 2)) / 2) * (n / 2);}}else if (cnt == 3) {if(!((n / 2) & 1)) {if(n % 3 == 0)ans = (n / 3) * (1ull + (n / 2)) * (n / 4);else if((1ull + (n / 2)) % 3 == 0){ans = n * ((1ull + (n / 2)) / 3) * (n / 4);}else if((n / 4) % 3 == 0){ans = n * (1ull + (n / 2)) * (n / 12);}}else {//ans = n * ((1ull + (n / 2)) / 2) * (n / 2)if(n % 3 == 0)ans = (n / 3) * ((1ull + (n / 2)) / 2) * (n / 2);else if((1ull + (n / 2)) % 3 == 0){ans = n * ((1ull + (n / 2)) / 6) * (n / 2);}else if((n / 4) % 3 == 0){ans = n * ((1ull + (n / 2)) / 2) * (n / 6);}}}}else {if(cnt == 1)ans = n * ((2ull + (n / 2)) * (n / 2 - 1)+ (n - 2));else if(cnt == 2) {if(!(n & 1)) {ans = (n / 2) * ((2ull + (n / 2)) * (n / 2 - 1)+ (n - 2));}else {ans = n * (((2ull + (n / 2)) * (n / 2 - 1) + (n - 2)) / 2);}}else if(cnt == 3) {if(!(n & 1)) {if((n / 2) % 3 == 0)ans = (n / 6) * ((2ull + (n / 2)) * (n / 2 - 1)+ (n - 2));elseans = (n / 2) * (((2ull + (n / 2)) * (n / 2 - 1)+ (n - 2)) / 3);}else {if(n % 3 == 0)ans = (n / 3) * (((2ull + (n / 2)) * (n / 2 - 1) + (n - 2)) / 2);elseans = n * ((((2ull + (n / 2)) * (n / 2 - 1) + (n - 2)) / 2) / 3);}}}cout << ans;return 0;
}

D - A Simple Stone Game

思维，其实感觉不是大模拟

在为什么要分解质因数这里刚开始脑子de牢想了一会

还好

#include <bits/stdC++.h>
using namespace std;typedef long long ll;const int maxn = 1e5 + 10;
ll a[maxn];
vector<ll> nums;
int n;ll getRes(ll x) {nums.clear();ll sum = 0;for(int i = 0; i < n; ++ i) {if(a[i] % x == 0) continue;ll num = a[i] % x;nums.push_back(num);sum += num;}sort(nums.begin(), nums.end());int siz = nums.size();ll res = 0;for(int i = siz - 1; i >= 0; -- i) {ll tmp = nums[i];res += x - tmp;sum -= x;if(sum <= 0) break;}return res;
}int main() {int T;cin >> T;while(T --) {//int n;cin >> n;ll sum = 0;ll mx = 0;for(int i = 0; i < n; ++ i) {cin >> a[i];sum += a[i];mx = max(mx, a[i]);}ll ans = sum - mx;//cout << sum << " " << mx << endl;for(ll i = 2; i <= sum / i; ++ i) {         if(sum % i == 0) {//cout << "i :" << i << endl;ll res = getRes(i);//cout << "----" << i << " " << res << endl;ans = min(ans, res);while(sum % i == 0) sum /= i;}}if(sum > 1) {ll res = getRes(sum);//cout << "----" << sum << " " << res << endl;ans = min(ans, res); }cout << ans << endl;}return 0;
}