暑假第十七测

题解：

第一题

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e5 + 10;
ll a[M], b[M], ans;
priority_queue <ll, vector<ll> , greater<ll> > Q;
int main(){freopen("buy.in","r",stdin);freopen("buy.out","w",stdout);int n;scanf("%d", &n);for(int i = 1; i <= n; i++)scanf("%I64d", &a[i]);for(int i = 1; i <= n; i++)scanf("%I64d", &b[i]);for(int i = 1; i <= n; i++){Q.push(a[i]);if(!Q.empty()){int u = Q.top();if(u < b[i]) {ans += b[i] - u;Q.pop();Q.push(b[i]);    }}}printf("%I64d\n",ans);
}

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第二题：

#include<bits/stdc++.h>
using namespace std;#define ll long long;
const int M = 1e5 + 10;
int n, a[M], sum[M];
struct Node{int mx, tag;Node *ls, * rs;void up(){mx = max(ls->mx, rs->mx);}void down(){if(tag){ls->tag += tag; rs->tag += tag;ls->mx += tag; rs->mx += tag;}tag = 0;}
}pool[M << 2], *tail = pool, *root;
Node * build(int lf = 1, int rg = n){Node * nd = ++tail;if(lf == rg);else {int mid = (lf + rg) >> 1;nd->ls = build(lf, mid);nd->rs = build(mid + 1, rg);}return nd;
}
#define Ls nd->ls, lf, mid
#define Rs nd->rs, mid+1, rg
void insert(int pos, int val, Node * nd = root, int lf = 1, int rg = n){if(lf == rg)nd->mx = val;else {int mid = (lf + rg) >> 1;if(pos <= mid)insert(pos, val, Ls);else insert(pos, val, Rs);nd->up();}
}
void modify(int L, int R, int val, Node * nd = root, int lf = 1, int rg = n){if(L <= lf && rg <= R){nd->mx += val;nd->tag += val;}else {nd->down();int mid = (lf + rg) >> 1;if(L <= mid)modify(L, R, val, Ls);if(R > mid) modify(L, R, val, Rs);nd->up();}
}
int query(int L, int R, Node * nd = root, int lf = 1, int rg = n){if(L <= lf && rg <= R)return nd->mx;else {nd->down();int mid = (lf + rg) >> 1;int ans = -2e9;if(L <= mid)ans = query(L, R, Ls);if(R > mid) ans = max(ans, query(L, R, Rs));return ans;}}int main(){freopen("invest.in","r",stdin);freopen("invest.out","w",stdout);int s, e;int ans = -2e9;scanf("%d%d%d", &n, &s, &e);root = build();for(int i = 1; i <= e; i++){scanf("%d", a + i);sum[i] = sum[i - 1] + a[i];insert(i, sum[i]);}int llf = 1, lf = s, rg = e, now = 1;ans = max(ans, query(lf, rg));//cout<<ans<<endl;for(int i = e+1; i <= n; i++){scanf("%d", a + i);sum[i] = sum[i - 1] + a[i];lf++, rg++;insert(rg, sum[i] - sum[i - e]);modify(llf+1, rg - 1, -a[now]);llf++;now++;int p = query(lf, rg);ans = max(ans, p);//cout<<ans<<endl;
    }for(int i = llf; i + s - 1 <= n; i++){modify(llf+1, n, -a[llf]);llf++;     ans = max(ans, query(i+s-1, n));}printf("%d\n",ans);return 0;
}

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#include<stdio.h>
int a[100002],k[100002];
int max(int x,int y)
{return x>y?x:y; 
}
int qin()
{char ch;int in=0;bool flag=0;while(ch!='-'&&!(ch>='0'&&ch<='9')) ch=getchar();if(ch=='-'){flag=1;ch=getchar();}do{in=in*10+ch-'0';ch=getchar();}while(ch>='0'&&ch<='9');if(flag)in*=-1;return in;
}
int co()
{freopen("invest.in","r",stdin);freopen("invest.out","w",stdout);int n,i,t,head,tail,s,ans=-10000000,j;scanf("%d%d%d",&n,&s,&t);for(i=1;i<=n;i++)a[i]=a[i-1]+qin();head=tail=1;for(i=s;i<=n;i++){while(head<tail&&i-k[head]>t) head++;while(head<tail&&a[i-s]<a[k[tail-1]]) tail--;//k[a[i]]?????k[tail++]=i-s;ans=max(ans,a[i]-a[k[head]]);}printf("%d\n",ans); 
}
int ccc=co();
int main(){;
}

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第三题：

#include<bits/stdc++.h>
using namespace std;
const int M = 1e6 + 10;
#define ll long long
const ll mod = 19930726;
char c[M], s[ M << 1 ];
ll cnt[M]; 
int len, pal[M << 1];
ll ksm(ll a, ll b){ll ret = 1;for( ; b; b >>= 1, a = a * a % mod){if(b & 1) ret = ret * a % mod;}return ret;
}void init(){len = strlen(c);s[0] = '$';for(int i = 0; i < len; i++){s[i * 2 + 1] = '#';s[i * 2 + 2] = c[i];}s[len * 2 + 1] ='#';s[len * 2 + 2] ='@';int L = len * 2 + 1;int id = 1, Maxid = 0;for(int i = 1; i <= L; i++){if(Maxid >= i)pal[i] = min(Maxid - i + 1, pal[2 * id - i]);else pal[i] = 1;while(s[i - pal[i]] == s[i + pal[i]])pal[i]++;if(i + pal[i] - 1 > Maxid) id = i, Maxid = i + pal[i] - 1;if((i&1) == 0)cnt[(pal[i] * 2 - 1) / 2]++;}}int main(){freopen("rehearse.in","r",stdin);freopen("rehearse.out","w",stdout);int n; ll k, ans = 1, sum = 0;cin>>n>>k;scanf("%s", c);init();int limit = n & 1 ? n : n - 1;for(int i = limit; i > 0; i -= 2){if(i != 1)cnt[i - 2] += cnt[i];if(sum + cnt[i] <= k)ans = (ans * ksm(i, cnt[i])) % mod;else ans = (ans * ksm(i, k - sum)) % mod;sum += cnt[i];if(sum >= k)break;} if(sum >= k)cout<<ans<<endl;else printf("-1\n");
}