题意好理解.
//A. Scarborough Fair 字符更换 #include<cstring> #include<cstdio> #include<algorithm> #include<cmath> #include<iostream> using namespace std; const int N=1e5+10; char s[N]; int main(){ int n,m; while(~scanf("%d%d",&n,&m)){ scanf("%s",s); int len=strlen(s); while(m--){ char a,b; int l,r; cin>>l>>r>>a>>b; for(int i=l-1;i<r;i++){ if(s[i]==a)s[i]=b; } } printf("%s ",s); } return 0; }
题意：前K个长度为偶数的回文数相加%p；
思路：长度为偶数，那么每个数对称一下都是符合要求的回文数
AC代码：
#include<iostream> #include<cstring> #include<algorithm> #include<cstdio> #include<queue> using namespace std; int k,p; long long zcy[100005]; void init() { int cnt=0; for(int i=1;i<=100000;i++) { long long tmp=i; int p=i; while(p){ tmp=tmp*10+p%10; p/=10; } zcy[++cnt]=tmp; } } int main() { init(); while(~scanf("%d%d",&k,&p)) { long long sum=0; for(int i=1;i<=k;i++){ sum+=zcy[i]; sum%=p; } printf("%lld ",sum); } return 0; }
递归，模拟
/* http://codeforces.com/contest/897/problem/C */ #include<bits/stdc++.h> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define LL long long using namespace std; inline LL read(); const int MAXN = 1e5+10; string s0 = "What are you doing at the end of the world? Are you busy? Will you save us?", s1 = "What are you doing while sending "",s2 = ""? Are you busy? Will you send "",s3 = ""?"; LL len[MAXN]; LL len0 = s0.size(),len1 = s1.size(),len2 = s2.size(),len3 = s3.size(); char fun(LL n,LL k) { if(!n) { if(k <= len0) return s0[k-1]; else return '.'; } if(k <= len1) return s1[k-1]; k -= len1; if(k <= len[n-1] || !len[n-1]) return fun(n-1,k); k -= len[n-1]; if(k <= len2) return s2[k-1]; k -= len2; if(k <= len[n-1] || !len[n-1]) return fun(n-1,k); k -= len[n-1]; if(k <= len3) return s3[k-1]; return '.'; } int main(void) { int q = read(); len[0] = len0; LL tmp = len1+len2+len3; for(int i = 1;len[i-1] <= 1e18; ++i) { len[i] = 2*len[i-1]+tmp; } while(q--) { LL n = read(),k = read(); putchar(fun(n,k)); } } inline LL read() { LL x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; }
交互
所以应该能想到从两边，开始扫描，让大于c/2的从n开始往下扫描，这样刚刚那组数据就可以过了，复杂度正好是n*c/2。
#include <bits/stdc++.h> using namespace std; int n; int a[1005]; int judge() { for(int i=1; i<=n; i++) { if(a[i]==0)return 0; if(i<n && a[i]>a[i+1])return 0; } return 1; } int main() { int m, i, j, c; cin>>n>>m>>c; for(i=1; i<=m; i++) { int x; scanf("%d", &x); if(x<=c/2) { for(int i=1; i<=n; i++) { if(a[i]==0 || a[i]>x) { a[i]=x; cout<<i<<endl; break; } } } else { for(int i=n; i>=1; i--) { if(a[i]==0 || a[i]<x) { a[i]=x; cout<<i<<endl; break; } } } if(judge())return 0; } return 0; }
E - Willem, Chtholly and Seniorious
珂朵莉树模板题。
以下给出些链接
#include<iostream> #include<cmath> #include<algorithm> #include<vector> #include<cstring> #include<cstdio> #include<set> using namespace std; typedef long long lt; #define IT set<node>::iterator #define pir pair<lt,int> #define mkp(x,y) make_pair(x,y) lt read() { lt f=1,x=0; char ss=getchar(); while(ss<'0'||ss>'9'){if(ss=='-')f=-1;ss=getchar();} while(ss>='0'&&ss<='9'){x=x*10+ss-'0';ss=getchar();} return f*x; } lt seed,vmax; lt rnd() { lt res=seed; seed=(seed*7+13)%1000000007; return res; } const int maxn=100010; int n,m; lt a[maxn]; struct node { int ll,rr; mutable lt val; node(int L,int R=-1,lt V=0): ll(L), rr(R), val(V) {} bool operator < (const node& tt)const { return ll<tt.ll;} }; set<node> st; lt qpow(lt a,lt k,lt p) { lt res=1; a%=p; while(k>0){ if(k&1) res=(res*a)%p; a=(a*a)%p; k>>=1; } return res; } IT split(int pos) { IT it=st.lower_bound(node(pos)); if(it!=st.end()&&it->ll==pos) return it; --it; int ll=it->ll,rr=it->rr; lt val=it->val; st.erase(it); st.insert(node(ll,pos-1,val)); return st.insert(node(pos,rr,val)).first; } void assign(int ll,int rr,lt val) { IT itr=split(rr+1),itl=split(ll); st.erase(itl,itr); st.insert(node(ll,rr,val)); } void add(int ll,int rr,lt val) { IT itr=split(rr+1),itl=split(ll); for(;itl!=itr;++itl) itl->val+=val; } lt kth(int ll,int rr,int k) { vector<pir> vec; IT itr=split(rr+1),itl=split(ll); for(;itl!=itr;++itl) vec.push_back(pir(itl->val,itl->rr-itl->ll+1)); sort(vec.begin(),vec.end()); for(vector<pir>::iterator it=vec.begin();it!=vec.end();++it) { k-=it->second; if(k<=0) return it->first; } return -1; } lt qsum(int ll,int rr,lt x,lt y) { lt res=0; IT itr=split(rr+1),itl=split(ll); for(;itl!=itr;++itl) res+=(qpow(itl->val,x,y)*((itl->rr-itl->ll+1)%y))%y,res%=y; return res; } int main() { n=read();m=read(); seed=read();vmax=read(); for(int i=1;i<=n;++i) { a[i]=(rnd()%vmax)+1; st.insert(node(i,i,a[i])); } for(int i=1;i<=m;++i) { int op=(rnd()%4)+1; int ll=(rnd()%n)+1,rr=(rnd()%n)+1; lt x,y; if(ll>rr) swap(ll,rr); if(op==3) x=(rnd()%(rr-ll+1))+1; else x=(rnd()%vmax)+1; if(op==4) y=(rnd()%vmax)+1; if(op==1) add(ll,rr,x); else if(op==2) assign(ll,rr,x); else if(op==3) printf("%lld ",kth(ll,rr,x)); else if(op==4) printf("%lld ",qsum(ll,rr,x,y)); } return 0; }
概率随机+set
2. 题意描述
有一个序列a[]a[],长度为nn。然后是mm次操作。
操作1：将al,al+1,…,aral,al+1,…,ar的数,全部加上xx;
操作2：将al,al+1,…,aral,al+1,…,ar的数,全部置位xx;
操作3：求al,al+1,…,aral,al+1,…,ar的数中的第xx小的数;
操作4：求(∑ri=laxi)%y(∑i=lraix)%y;
注意：序列a[]a[]，以及所有操作（即op,l,r,x,yop,l,r,x,y）都是等概率的随机生成的。
数据范围：
1 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 1091 ≤ n, m ≤ 105, 0 ≤ seed < 109 + 7, 1 ≤ vmax ≤ 109
3. 解题思路
上面四种操作，出现的概率都是1414。
假如用线段来描述序列a[]a[]中值相同的子序列。当n=105n=105时，初始的时候，序列a[]a[]随机分布，所以线段很很多。但是经过多次操作2之后，会让这个序列极速收敛，最终收敛到小于5050左右。然而概率论太差，我并不会证明合并到线段数目小于某一个值的次数的数学期望。
但是自己打印了一下，大概250250次合并就能让线段数目小于200。
然后，知道了这种性质之后，就可以用setset来保存维护线段。剩下的操作就是暴力删除一堆线段，暴力插入一些线段，暴力对线段进行查询。
#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int inf = 0x3f3f3f3f; const ll infl = 0x3f3f3f3f3f3f3f3fLL; template<typename T> inline void umax(T &a, T b) { a = max(a, b); } template<typename T> inline void umin(T &a, T b) { a = min(a, b); } void debug() { cout << endl; } template<typename T, typename ...R> void debug (T f, R ...r) { cout << "[" << f << "]"; debug (r...); } const int MAXN = 100005; ll n, m, seed, vmax, a[MAXN]; ll rnd() { ll ret = seed; seed = (seed * 7 + 13) % 1000000007; return ret; } struct Line { pii itv; ll val; Line() {} Line(pii itv, ll val) : itv(itv), val(val) {} bool operator < (const Line& line) const { return itv < line.itv; } }; bool cmp_val(const Line& l1, const Line& l2) { return l1.val < l2.val; } set<Line> st; set<Line>::iterator it, it1, it2; void oper1(int l, int r, ll x) { // [it1, it2) it1 = st.upper_bound(Line(pii(l, inf), 0ll)); --it1; it2 = st.upper_bound(Line(pii(r, inf), 0ll)); vector<Line> buf; for (it = it1; it != it2; ++it) { if ((it->itv).first > r) break; buf.push_back(*it); } st.erase(it1, it2); if (buf[0].itv.first < l) { st.insert(Line(pii(buf[0].itv.first, l - 1), buf[0].val)); buf[0].itv.first = l; } int sz = buf.size(); if (buf[sz - 1].itv.second > r) { st.insert(Line(pii(r + 1, buf[sz - 1].itv.second), buf[sz - 1].val)); buf[sz - 1].itv.second = r; } for (Line& line : buf) { line.val += x; st.insert(line); } // for(int i = l; i <= r; ++i) a[i] += x; } void oper2(int l, int r, ll x) { // [it1, it2) it1 = st.upper_bound(Line(pii(l, inf), 0ll)); --it1; it2 = st.upper_bound(Line(pii(r, inf), 0ll)); vector<Line> buf; for (it = it1; it != it2; ++it) { if ((it->itv).first > r) break; buf.push_back(*it); } st.erase(it1, it2); if (buf[0].itv.first < l && buf[0].val != x) { st.insert(Line(pii(buf[0].itv.first, l - 1), buf[0].val)); buf[0].itv.first = l; } int sz = buf.size(); if (buf[sz - 1].itv.second > r && buf[sz - 1].val != x) { st.insert(Line(pii(r + 1, buf[sz - 1].itv.second), buf[sz - 1].val)); buf[sz - 1].itv.second = r; } st.insert(Line(pii(buf[0].itv.first, buf[sz - 1].itv.second), x)); // for(int i = l; i <= r; ++i) a[i] = x; } ll oper3(int l, int r, ll x) { // [it1, it2) it1 = st.upper_bound(Line(pii(l, inf), 0ll)); --it1; it2 = st.upper_bound(Line(pii(r, inf), 0ll)); vector<Line> buf; for (it = it1; it != it2; ++it) { if ((it->itv).first > r) break; buf.push_back(*it); } if (buf[0].itv.first < l) { buf[0].itv.first = l; } int sz = buf.size(); if (buf[sz - 1].itv.second > r) { buf[sz - 1].itv.second = r; } sort(buf.begin(), buf.end(), cmp_val); ll w = 0; for (Line& line : buf) { w += (line.itv.second - line.itv.first + 1); if (w >= x) return line.val; } return -1; } ll qpow(ll a, ll b, ll mod) { ll ret = 1; a = a % mod; // notice here!!! while (b > 0) { if (b & 1) ret = ret * a % mod; a = a * a % mod; b >>= 1; } return ret; } ll oper4(int l, int r, ll x, ll y) { // [it1, it2) it1 = st.upper_bound(Line(pii(l, inf), 0ll)); --it1; it2 = st.upper_bound(Line(pii(r, inf), 0ll)); vector<Line> buf; for (it = it1; it != it2; ++it) { if ((it->itv).first > r) break; buf.push_back(*it); } if (buf[0].itv.first < l) { buf[0].itv.first = l; } int sz = buf.size(); if (buf[sz - 1].itv.second > r) { buf[sz - 1].itv.second = r; } ll ret = 0; for (Line& line : buf) { int w = (line.itv.second - line.itv.first + 1); ret += (ll)w * qpow(line.val, x, y) % y; ret %= y; } return ret; } int main() { #ifdef ___LOCAL_WONZY___ freopen("input.txt", "r", stdin); freopen("output.txt", "w+", stdout); #endif // ___LOCAL_WONZY___ cin >> n >> m >> seed >> vmax; for (int i = 1; i <= n; ++i) a[i] = (rnd() % vmax) + 1; st.clear(); for (int i = 1; i <= n; ++i) { int j = i; while (j + 1 <= n && a[i] == a[j + 1]) ++j; st.insert(Line(pii(i, j), a[i])); i = j; } int op, l, r; ll x, y; int cnt = 0; for (int i = 1; i <= m; ++i) { op = rnd() % 4 + 1; l = rnd() % n + 1; r = rnd() % n + 1; if (l > r) swap(l, r); if (op == 3) x = (rnd() % (r - l + 1)) + 1; else x = (rnd() % vmax) + 1; if (op == 4) y = (rnd() % vmax) + 1; // debug(op, l, r, x, y); if (op == 1) oper1(l, r, x); else if (op == 2) oper2(l, r, x); else if (op == 3) { ll ret = oper3(l, r, x); cout << ret << endl; } else { ll ret = oper4(l, r, x, y); cout << ret << endl; } } #ifdef ___LOCAL_WONZY___ cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC * 1000 << "ms." << endl; #endif // ___LOCAL_WONZY___ return 0; }