(A. Level Statistics)
每天都可能会有人玩游戏,同时一部分人会过关,玩游戏的人数和过关的人数会每天更新,问记录的数据是否没有矛盾
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
void solve(){
int n, x = 0, y = 0;
cin >> n;
bool ok = true;
for(int i = 1; i <= n; i++){
int a, b;
cin >> a >> b;
if(a<x or b<y or b>a) ok = false;
if(b-y>a-x) ok = false;
x = a, y = b;
}
if(ok) cout << "YES" << endl;
else cout << "NO" << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
(B - Middle Class)
每次可以选几个人把钱均摊,问经过若干次操作之后最多有多少人的钱可以大于(x)
从大到小排序,然后判断前(i)个人均摊之后能否都大于(x),找最大的(i)即可
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e5+7;
using LL = int_fast64_t;
int n,x,A[MAXN];
void solve(){
cin >> n >> x;
for(int i = 1; i <= n; i++) cin >> A[i];
sort(A+1,A+1+n,greater<int>());
LL tot = 0, pr = 0;
for(int i = 1; i <= n; i++){
pr += x;
tot += A[i];
if(tot<pr){
cout << i-1 << endl;
return;
}
}
cout << n << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
(C. Circle of Monsters)
怪物站成一圈,每个怪物有血量(A_i),和爆炸伤害(B_i),每次打一颗子弹,可以扣除怪物(1)滴血,怪物血量(le0)时死亡,死亡之后会对(i+1)个怪物造成相应的爆炸伤害,问最少多少子弹可以消灭所有怪物。
(1~n)复制一遍到(n+1~2n),然后算上一个怪物爆炸的情况下这个怪物还需要多少子弹,求前缀和,然后枚举第一个打的怪物即可
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 6e5+7;
using LL = int_fast64_t;
int n;
LL A[MAXN],B[MAXN],cost[MAXN],sum[MAXN];
void solve(){
cin >> n;
for(int i = 1; i <= n; i++){
cin >> A[i] >> B[i];
A[i+n] = A[i]; B[i+n] = B[i];
}
for(int i = 2; i <= n * 2; i++){
cost[i] = max(0ll,A[i]-B[i-1]);
sum[i] = sum[i-1] + cost[i];
}
LL ret = 0x3f3f3f3f3f3f3f3f;
for(int i = 1; i <= n; i++) ret = min(ret,A[i]+sum[i+n-1]-sum[i]);
cout << ret << endl;
}
int main(){
____();
int T;
for(cin >> T; T; T--) solve();
return 0;
}
(D. Minimum Euler Cycle)
给一张完全图,任意两个点(u,v)之间有(u
ightarrow v, v
ightarrow u)两条边,要求走完所有边并且走的点的标号的字典序最小,要求输出一段区间内的顶点标号。
假设为顶点数为(n)的完全图,则方案为:
((1, 2, 1, 3, 1, 4 cdots 1, n)( 2, 3, 2, 4 cdots 2, n)( 3, 4, 3, 5 cdots 3, n) cdots (n-3, n-2, n-3, n-1, n-3, n)( n-2, n-1, n-2, n)( n-1, n) 1)
用括号括出来的可以找出来规律
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
LL n, l, r;
void solve(){
scanf("%I64d %I64d %I64d",&n,&l,&r);
if(l==n*(n-1)+1){
puts("1");
return;
}
int stpos = 1;
LL tot = 0;
LL L = 0, R = n - 1;
while(L<=R){
LL mid = (L+R)>>1;
if((n-1+n-mid)*mid<l) L = mid + 1;
else R = mid - 1;
}
stpos = L;
tot = (n-1+n-R)*R;
int cur = stpos + (l-tot) / 2;
bool tg = (((l-tot)&1)?true:false);
for(LL i = l; i <= r; i++){
if(stpos==n){
printf("1 ");
break;
}
if(tg){
printf("%d ",stpos);
cur++;
}
else{
printf("%d ",cur);
if(cur==n){
stpos++;
cur = stpos;
}
}
tg = !tg;
}
puts("");
}
int main(){
int T;
for(scanf("%d",&T); T; T--) solve();
return 0;
}
(E. Divisor Paths)
给出数(D),以这个数建图,建图规则为:所有(D)的因子为顶点,(u,v(u>v))两点之间连边的条件为:(u%v==0 and u/v is prime),且边权为(f(u)-f(v)),其中(f(x))为(x)的因子数量,现在有(Q)次询问,每次询问(u,v)间的最短路条数。
首先我们假设要从(u)到(v)且(u\%v==0),那么走的边必然是(u)比(v)分解之后多的质因子,略证:
首先假设命题成立
现在从(a
ightarrow b),我们走多余质因子边:(path(a,b)=a
ightarrow c
ightarrow b),则路径长度为:(f(a)-f(c)+f(c)-f(b)=f(a)-f(b)),现在假设不走多余质因子边则:(path(a,b)=a
ightarrow c
ightarrow d
ightarrow cdots
ightarrow b),路径长度为(f(a)-f(c)+f(d)-f(c)+dist(d,b)=f(a)-f(c)+f(d)-f(c)+f(d)-f(x)+dist(x,b)=dist(a,b)+2f(d)-2f(c)>dist(a,b))
如果(u\%v
e 0),则必然是先从(u)到(gcd(u,v)),再从(gcd(u,v)到v)是最优的
如果两个数不互质的话可以先取(gcd),然后必然能经过点(1),然后只要找出(1)到各个点最短路的方案数即可。
假设(x)各个素因子的数量分别为(P_1,P_2,cdots,P_m),则最短路数量就是(frac{(sum_{i=1}^{m}P_i)!}{prod_{i=1}^{m}(P_i!)})
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
using LL = int_fast64_t;
const LL MOD = 998244353;
const int MAXN = 1e5+7;
LL D, q, fact[MAXN], invf[MAXN];
map<LL,LL> ret;
LL qpow(LL a, LL b){
LL res = 1;
while(b){
if(b&1) res = res * a % MOD;
b >>= 1;
a = a * a % MOD;
}
return res;
}
void ikun(LL x){
LL v = x;
vector<int> vec;
int tot = 0;
for(LL i = 2; i * i <= x; i++){
if(x%i!=0) continue;
int cnt = 0;
while(x%i==0) cnt++, x /= i;
tot += cnt;
vec.emplace_back(cnt);
if(x==1) break;
}
if(x!=1) tot++;
LL res = fact[tot];
for(int p : vec) res = res * invf[p] % MOD;
ret.insert(make_pair(v,res));
}
void solve(){
fact[0] = 1;
for(int i = 1; i < MAXN; i++) fact[i] = fact[i-1] * i % MOD;
invf[MAXN-1] = qpow(fact[MAXN-1],MOD-2);
for(int i = MAXN - 2; i >= 0; i--) invf[i] = invf[i+1] * (i+1) % MOD;
cin >> D >> q;
for(LL i = 1; i * i <= D; i++){
if(D%i!=0) continue;
ikun(i);
if(i*i!=D) ikun(D/i);
}
while(q--){
LL u, v;
cin >> u >> v;
LL g = __gcd(u,v);
u /= g; v /= g;
cout << ret.at(u) * ret.at(v) % MOD << endl;
}
}
int main(){
____();
solve();
return 0;
}
(F. Strange Function)
给出一个序列,(f(A))操作就是取序列(A)的前缀递增序列,序列中每个数有一个价值,现在要求你删除序列(a)中的一些元素,使得(f(A)=B)
考虑朴素的(DP),(dp[i][j])表示现在(A)串的(i)位置和(B)串的(j)位置匹配,最小的花费,转移如下:
①如果不删除
(egin{cases} dp[i+1][j] = dp[i][j] & if A[i+1]le B[j] \ dp[i+1][j+1] = dp[i][j] & if A[i+1]=B[j+1]
\ 不合法 & if A[i+1]>B[j] and A[i+1]
e B[j+1] end{cases})
②如果删除
(dp[i+1][j] = dp[i][j] + cost[i+1])
现在整理一下:
(egin{cases}dp[i+1][j] = dp[i][j]+min(cost[i+1],0) & if A[i+1]le B[j] \ dp[i+1][j+1] = min(dp[i][j],dp[i][j+1]) & if A[i+1]=B[j+1] \ dp[i+1][j] = dp[i][j]+cost[i+1] & if A[i+1]>B[j] end{cases})
可以发现(i+1)的状态都是从(i)的状态转移过来的,所以可以把第一维给去掉,然后用线段树或者树状数组维护一下
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 5e5+7;
using LL = int_fast64_t;
const LL INF = 0x3f3f3f3f3f3f3f3f;
int n,m,A[MAXN],B[MAXN],p[MAXN];
template<typename T>
struct SegmentTree{
T lazy[MAXN<<2];
int l[MAXN<<2],r[MAXN<<2];
#define ls(rt) rt << 1
#define rs(rt) rt << 1 | 1
void build(int L, int R, int rt){
l[rt] = L; r[rt] = R;
lazy[rt] = 0;
if(L+1==R) return;
int mid = (L + R) >> 1;
build(L,mid,ls(rt)); build(mid,R,rs(rt));
}
void pushdown(int rt){
if(!lazy[rt]) return;
lazy[ls(rt)] += lazy[rt];
lazy[rs(rt)] += lazy[rt];
lazy[rt] = 0;
}
void update(int L, int R, int rt, T x){
if(L>=r[rt] || l[rt]>=R) return;
if(L<=l[rt] && r[rt]<=R){
lazy[rt] += x;
return;
}
pushdown(rt);
update(L,R,ls(rt),x); update(L,R,rs(rt),x);
}
T query(int L, int R, int rt){
if(L>=r[rt] || l[rt]>=R) return 0;
if(L<=l[rt] && r[rt]<=R) return lazy[rt];
pushdown(rt);
return query(L,R,ls(rt))+query(L,R,rs(rt));
}
};
SegmentTree<LL> ST;
int main(){
____();
cin >> n;
for(int i = 1; i <= n; i++) cin >> A[i];
for(int i = 1; i <= n; i++) cin >> p[i];
cin >> m;
for(int i = 1; i <= m; i++) cin >> B[i];
ST.build(0,m+1,1);
ST.update(1,m+1,1,INF);
for(int i = 1; i <= n; i++){
int pos = lower_bound(B+1,B+1+m,A[i]) - B;
if(A[i]==B[pos]){
ST.update(pos,m+1,1,min(0,p[i]));
ST.update(pos,pos+1,1,min(0ll,ST.query(pos-1,pos,1)-ST.query(pos,pos+1,1)));
ST.update(0,pos,1,p[i]);
}
else{
ST.update(pos,m+1,1,min(0,p[i]));
ST.update(0,pos,1,p[i]);
}
}
LL ret = ST.query(m,m+1,1);
if(ret>INF/2) cout << "NO" << endl;
else cout << "YES" << endl << ret << endl;
return 0;
}
(G. Substring Search)
给出两个串(s,t)要求找到在(t)串中所有能和(s)串匹配的位置,其中字符匹配的条件是(s_j==t_{i+j} or P_{s_j}=t_{i+j}),其中(P)为(0-25)的一个排列。
如果用朴素的想法,(26)个字符每个都匹配一遍,最后找每个位置的匹配数量的话,一共要进行(78)遍(FFT),但是题目只给了(1.25s),显然是行不通的
如果转化成方程的形式,那么如果两个串匹配,则每个相应字符都能匹配上,可以写成:
把(s)串翻转后得到:
转换成卷积形式:
每个位置的值即为:
展开:
对于第一项和最后一项可以用前缀和来求,这样可以只做(7)次(FFT)
hack数据有点恶心,把NTT的998244353模数卡了,用1004535809能过
好像bitset可以水过,而且耗时还很优秀
//#pragma GCC optimize("O3")
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<bits/stdc++.h>
using namespace std;
function<void(void)> ____ = [](){ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);};
const int MAXN = 1e6+7;
using LL = int_fast64_t;
//const LL MOD = 998244353;
const LL MOD = 1004535809;
const LL g = 3;
int n,P[30],l,limit,r[MAXN],lens,lent;
LL A[MAXN],B[MAXN],C[MAXN];
char s[MAXN],t[MAXN];
LL qpow(LL a, LL b){
if(!a) return 0;
LL ret = 1;
while(b){
if(b&1) ret = ret * a % MOD;
b >>= 1;
a = a * a % MOD;
}
return ret;
}
void NTT(LL arr[], int rev){
for(int i = 0; i < limit; i++) if(i<r[i]) swap(arr[i],arr[r[i]]);
for(int len = 1; len < limit; len <<= 1){
LL wn = qpow(g,(MOD-1)/(len<<1));
if(rev==-1) wn = qpow(wn,MOD-2);
for(int R = 0; R < limit; R += (len<<1)){
LL w = 1;
for(int i = R; i < R + len; i++){
LL x = arr[i], y = w * arr[i+len] % MOD;
arr[i] = (x+y)%MOD;
arr[i+len] = (x-y+MOD)%MOD;
w = w * wn % MOD;
}
}
}
}
int main(){
for(int i = 0; i < 26; i++){
scanf("%d",&P[i]);
P[i]--;
}
scanf("%s %s",s,t);
lens = strlen(s); lent = strlen(t);
reverse(s,s+lens);
limit = 1; l = 0;
while(limit<=(lens+lent-2)) limit <<= 1, l++;
for(int i = 0; i < limit; i++) r[i] = (r[i>>1]>>1) | ((i&1)<<(l-1));
for(int i = lent; i < limit; i++) B[i] = 0;
for(int i = lens; i < limit; i++) C[i] = 0;
for(int i = 0; i < lent; i++){
LL x = t[i] - 'a';
B[i] = x * x * x;
}
for(int i = 0; i < lens; i++) C[i] = (-2*(s[i]-'a'+P[s[i]-'a'])+MOD)%MOD;
NTT(B,1); NTT(C,1);
for(int i = 0; i < limit; i++) A[i] = (A[i] + B[i] * C[i]) % MOD;
for(int i = lent; i < limit; i++) B[i] = 0;
for(int i = lens; i < limit; i++) C[i] = 0;
for(int i = 0; i < lent; i++){
LL x = t[i] - 'a';
B[i] = x * x;
}
for(int i = 0; i < lens; i++) C[i] = qpow(s[i]-'a',2)+qpow(P[s[i]-'a'],2)+4*(s[i]-'a')*P[s[i]-'a'];
NTT(B,1); NTT(C,1);
for(int i = 0; i < limit; i++) A[i] = (A[i] + B[i] * C[i]) % MOD;
for(int i = lent; i < limit; i++) B[i] = 0;
for(int i = lens; i < limit; i++) C[i] = 0;
for(int i = 0; i < lent; i++) B[i] = t[i] - 'a';
for(int i = 0; i < lens; i++) C[i] = (-2*(s[i]-'a')*P[s[i]-'a']*(s[i]-'a'+P[s[i]-'a'])+MOD)%MOD;
NTT(B,1); NTT(C,1);
for(int i = 0; i < limit; i++) A[i] = (A[i] + B[i] * C[i]) % MOD;
NTT(A,-1);
LL inv = qpow(limit,MOD-2);
LL sum = 0;
vector<LL> pre(lent);
for(int i = 0; i < lens; i++) sum += (s[i]-'a')*(s[i]-'a')*P[s[i]-'a']*P[s[i]-'a'];
pre[0] = qpow(t[0]-'a',4);
for(int i = 1; i < lent; i++) pre[i] = pre[i-1] + qpow(t[i]-'a',4);
for(int i = lens - 1; i < lent; i++) A[i] = (A[i] * inv + sum + pre[i] - ((i-lens>=0)?pre[i-lens]:0)) % MOD;
for(int i = lens - 1; i < lent; i++) putchar(A[i]?'0':'1'); puts("");
return 0;
}