emmm........很久很久以前把6.2过了所以emmmmmm 直接跳过，从6.1到6.3吧

Fence Rails

题目大意

N<=50个数A1,A2...

1023个数,每个数数值<=128,B

问 A 们能拆分成多少个B,求最多的个数

样例解释

A:
30=30
40=18+19+3
50=15+16+17+2
25=24
B:
15 (ok)
16 (ok)
17 (ok)
18 (ok)
19 (ok)
20
21
25
24 (ok)
30 (ok)

所以最多7个

题解

首先如果对B 排序，假设最优个数为k个

显然如果k个可行，那么排序后的B 的前k个可行

又如果k个可行那么k-1个可行，综上又满足二分

先 sort+二分+从大到小放b （第4个点就超时了）

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)using namespace std;const string filename = "fence8";void usefile(){freopen((filename+".in").c_str(),"r",stdin);freopen((filename+".out").c_str(),"w",stdout);
}int n;
int a[100];
int la[100];
int suma;
int R;
int r[1100];
int sumr[1100];int dfs(int idx){per(i,0,n){if(a[i] < r[idx]){return false;}if(a[i] ==  a[i+1] && la[i] == la[i+1]){continue;}if(la[i] < r[idx]){continue;}if(idx == 0){return true;}la[i] -= r[idx];int ret = dfs(idx-1);la[i] += r[idx];if(ret){return true;}}return false;
}bool test(int idx){if(sumr[idx] > suma)return false;return dfs(idx);
}int main(){usefile();scanf("%d",&n);rep(i,0,n){scanf("%d",a+i);}rep(i,0,n){la[i]=a[i];suma += a[i];}sort(a,a+n);scanf("%d",&R);rep(i,0,R){scanf("%d",r+i);}sort(r,r+R);if(r[0] > a[n-1]){cout<<0<<endl;return 0;}sumr[0]=r[0];rep(i,1,R){sumr[i]=sumr[i-1]+r[i];}int l=0,r=R;while(l+1<r){int mid = (l+r)/2;if(test(mid)){l = mid;}else{r = mid;}}cout<<l+1<<endl;return 0;
}

对B 的枚举过程加了相同长度的枚举优化 tle 5

int dfs(int idx,int stn = n){per(i,0,stn){if(a[i] < r[idx]){return false;}if(a[i] ==  a[i+1] && la[i] == la[i+1]){continue;}if(la[i] < r[idx]){continue;}if(idx == 0){return true;}la[i] -= r[idx];int ret;if(r[idx] == r[idx+1]){ret = dfs(idx-1,i+1);}else{ret = dfs(idx-1);}la[i] += r[idx];if(ret){return true;}}return false;
}

增加了无效残余木料 AC

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)using namespace std;const string filename = "fence8";void usefile(){freopen((filename+".in").c_str(),"r",stdin);freopen((filename+".out").c_str(),"w",stdout);
}int n;
int a[100];
int la[100];
int suma;
int R;
int r[1100];
int sumr[1100];int dfs(int idx,int stn = n){if(suma < sumr[idx]){return false;}per(i,0,stn){if(a[i] < r[idx]){return false;}if(a[i] ==  a[i+1] && la[i] == la[i+1]){continue;}if(la[i] < r[idx]){continue;}if(idx == 0){return true;}la[i] -= r[idx];suma-=r[idx];bool predel = la[i] < r[0];if(predel){suma -= la[i];}int ret;if(idx > 0 && r[idx-1] == r[idx]){ret = dfs(idx-1,i+1);}else{ret = dfs(idx-1);}if(predel){suma += la[i];}suma+=r[idx];la[i] += r[idx];if(ret){return true;}}return false;
}bool test(int idx){if(sumr[idx] > suma)return false;return dfs(idx);
}int main(){usefile();scanf("%d",&n);rep(i,0,n){scanf("%d",a+i);}rep(i,0,n){la[i]=a[i];suma += a[i];}sort(a,a+n);scanf("%d",&R);rep(i,0,R){scanf("%d",r+i);}sort(r,r+R);if(r[0] > a[n-1]){cout<<0<<endl;return 0;}sumr[0]=r[0];rep(i,1,R){sumr[i]=sumr[i-1]+r[i];}int l=0,r=R;while(l+1<r){int mid = (l+r)/2;if(test(mid)){l = mid;}else{r = mid;}}cout<<l+1<<endl;return 0;
}

综上二分+暴搜+减枝+处理顺序贪心

Cryptcowgraphy

题目大意

一个字符串能否通过正数次操作使得变为

Begin the Escape execution at the Break of Dawn

一次操作: 选取 ...C...O...W...,把C->O的字符串和O->W的字符串交换，然后去掉这三个选中C,O,W

题解

~~显然特判打表~~

我们已经知道目标串和起始串

所以如果可行，那么个数关系有C=O=W =(len(起始串)-len(目标串))/3,所以预先筛选的一个优化是，统计各个字符的个数，和目标串进行比较

所以当比较是可能时，答案要么0 0,要么 1 字母C 的个数

我们可以优化的点

字母个数统计
被C O W 分割的在任意时刻是目标串的子串
搜索顺序先O
字符串hash [注意这个方法如果你是在打cf，那么很可能被hack

注意输入是一行....所以不要scanf %s

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)using namespace std;const string filename = "cryptcow";void usefile(){freopen((filename+".in").c_str(),"r",stdin);freopen((filename+".out").c_str(),"w",stdout);
}
const char Goal[] = "Begin the Escape execution at the Break of Dawn";
const int Mod = 999983;char s[110];
int ans, cnt;
bool hsh[Mod];// 删除a b c位置上的, 交换a->b b->c
void work(int a, int b, int c) {static char tmp[100];int len = strlen(s), tot = 0;for(int i = 0; i < a; ++i) {tmp[tot++] = s[i];}for(int i = b + 1; i < c; ++i) {tmp[tot++] = s[i];}for(int i = a + 1; i < b; ++i) {tmp[tot++] = s[i];}for(int i = c + 1; i < len; ++i) {tmp[tot++] = s[i];}tmp[tot] = 0;strcpy(s, tmp);
}int getHash() {int ret = 0, len = strlen(s);for(int i = 0; i < len; ++i) {int num = (s[i]==' ')?1:(isupper(s[i]) ? s[i] - 'A' + 2 : s[i] - 'a' + 28);ret = (ret * 57 + num) % Mod;}return ret;
}bool dfs(int depth) {if(strcmp(s, Goal) == 0) {ans = depth;return true;}int x = getHash();if(hsh[x]) {return false;}hsh[x] = true;++cnt;// 被C O W 分割的 字串应该是Goal的连续子串static char sub[100];int len = strlen(s);int c[20], o[20], w[20];c[0] = o[0] = w[0] = 0;for(int i = 0, j = 0; i < len; ++i) {if(s[i] == 'C' || s[i] == 'O' || s[i] == 'W') {if(s[i] == 'C') {c[++c[0]] = i;}if(s[i] == 'O') {o[++o[0]] = i;}if(s[i] == 'W') {w[++w[0]] = i;}sub[j] = 0;if(!strstr(Goal, sub)) { //return false;}j = 0;}else {sub[j++] = s[i];}}// C = W = Oif(o[0] != c[0] || o[0] != w[0] || w[0] != c[0]) {return false;}char pre[100];strcpy(pre, s); // 递归暂存// 查找顺序 先找Orep(j,1,o[0]+1){per(k,1,w[0]+1){if(w[k] < o[j])break;rep(i,1,c[0]+1){if(c[i] > o[j])break;work(c[i], o[j], w[k]);bool ret = dfs(depth + 1);if(ret){return true;}if(cnt > 200000) { // ............................return false;}strcpy(s, pre);}}}return false;
}int main() {usefile();cin.getline(s,100);// scanf("%s",s);int ret = dfs(0);printf("%d %d\n", ret, ans);return 0;
}

Cowcycles

题目大意

25 <= F1 < F2 <= 80

在[F1,F2]范围找[1,5]个数f1,f2..

5 <= R1 < R2 <= 40

在[R1,R2]范围找[1,10]个数r1,r2..

ratio(i,j) = fi/rj

在max ratio/min ratio >= 3的限制下

把所有ratio(i,j)排序，最小化排序后数组相邻值的差的方差

求具体方案

题解

ratio(i1,j1)/ratio(i2,j2) = i1*j2/i2*j1

要使这个值的最大值大于3,注意到都是正数，也就是(max(i)*max(j))/(min(i)*min(j)) >= 3

然后因为ratio要先sort，再最小化差的方差，就感觉无路可推，只能暴搜

优化

默认的限制减枝
个数少，运算过程相对有序 --> 计算顺序+插入排序
搜一搜初中的方差变形公式，省掉最外层的1/n 有只用比较sum{平方}+(sum{}平方)/n

注意以下代码过了USACO 但是有潜在风险，浮点数比大小！！ 如果是打cf，可能会被叉

#include <bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)using namespace std;const string filename = "cowcycle";void usefile(){freopen((filename+".in").c_str(),"r",stdin);freopen((filename+".out").c_str(),"w",stdout);
}int s1[20],s2[20];
int ans1[20],ans2[20];
int F,F1,F2;
int R,R1,R2;
int cnt;
double rate[100],diff[100];
double minvf=10000000;void count(){int k=0;double sum=0,avg,vf=0,sumf=0,p;// 数据量小 采用插入排序rep(i,0,F){rep(j,0,R){p=(double)s1[i]/s2[j];int l=++k;while (rate[l-1]>=p) {rate[l]=rate[l-1];l--;}rate[l]=p;}}rep(i,1,cnt){diff[i]=rate[i+1]-rate[i];sum+=diff[i];sumf+=diff[i]*diff[i];}avg=sum/(cnt-1);// 相邻值的差的个数 比值的个数少1vf=sumf-sum*avg;if (vf<minvf)  {minvf=vf;memcpy(ans1,s1,sizeof(int)*F);memcpy(ans2,s2,sizeof(int)*R);}
}// 枚举后齿轮 从w 到R2-R+k+1
void sc2(int k,int w){if (k==R){if (s1[F-1]*s2[R-1]<3*s1[0]*s2[0]) // 题目限制条件剪枝return;count();return;}rep(i,w,R2-R+k+2){s2[k]=i;sc2(k+1,i+1);}
}// 枚举前齿轮 从w到F2-F+k+1
void sc1(int k,int w){if (k==F)  {sc2(0,R1);return;}rep(i,w,F2-F+k+2){s1[k]=i;sc1(k+1,i+1);}
}int main() {usefile();cin>> F >> R >> F1 >> F2 >> R1 >> R2;cnt=F*R;sc1(0,F1);rep(i,0,F){cout << ans1[i] << " \n"[i==F-1];}rep(i,0,R){cout << ans2[i] << " \n"[i==R-1];}return 0;
}