水题略过
#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;
typedef long long int LL;
int main()
{
LL a,b;
scanf("%I64d%I64d",&a,&b);
LL num=1;
while(true)
{
if(a<num){printf("Vladik\n");return 0;}
a-=num;
//b+=num;
num++;
if(b<num){printf("Valera\n");return 0;}
b-=num;
//a+=num;
num++;
}
return 0;
}
B. Vladik and Complicated Book
给定一个序列, 询问子区间【l,r】的第k小数是不是原位置的数。
显然可以用主席树维护,不过这道题数据放水了,用暴力一点的方法应该也能过。
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 100000 + 5;
int a[N], b[N], rt[N * 20], ls[N * 20], rs[N * 20], sum[N * 20],numm[N];
int n, k, tot, sz, ql, qr, x, q, T;
void Build(int& o, int l, int r){
if(l>r)return ;
o = ++ tot;
sum[o] = 0;
if(l == r) return;
int m = (l + r) >> 1;
Build(ls[o], l, m);
Build(rs[o], m + 1, r);
}
void update(int& o, int l, int r, int last, int p){
o = ++ tot;
ls[o] = ls[last];
rs[o] = rs[last];
sum[o] = sum[last] + 1;
if(l == r) return;
int m = (l + r) >> 1;
if(p <= b[m]) update(ls[o], l, m, ls[last], p);
else update(rs[o], m + 1, r, rs[last], p);
}
int query(int ss, int tt, int l, int r, int k){
if(l == r) return l;
int m = (l + r) >> 1;
int cnt = sum[ls[tt]] - sum[ls[ss]];
if(k <= cnt) return query(ls[ss], ls[tt], l, m, k);
else return query(rs[ss], rs[tt], m + 1, r, k - cnt);
}
void work(){
scanf("%d%d%d", &ql, &qr, &x);
int xx=x;
x=(xx-ql+1);
int ans = query(rt[ql - 1], rt[qr], 1, sz, x);
if(b[ans]==numm[xx])printf("Yes\n");
else printf("No\n");
}
int main(){//freopen("t.txt","r",stdin);
T=1;
while(T--){
scanf("%d%d", &n, &q);
for(int i = 1; i <= n; i ++) scanf("%d", a + i), numm[i]=b[i] = a[i];
sort(b + 1, b + n + 1);
sz = unique(b + 1, b + n + 1) - (b + 1);
tot = 0;
Build(rt[0],1, sz);
//for(int i = 0; i <= 4 * n; i ++)printf("%d,rt = %d,ls = %d, rs = %d, sum = %d\n", i, rt[i], ls[i], rs[i], sum[i]);
//for(int i = 1; i <= n; i ++)a[i] = lower_bound(b + 1, b + sz + 1, a[i]) - b;
for(int i = 1; i <= n; i ++)update(rt[i], 1, sz, rt[i - 1], a[i]);
// for(int i = 0; i <= 5 * n; i ++)printf("%d,rt = %d,ls = %d, rs = %d, sum = %d\n", i, rt[i], ls[i], rs[i], sum[i]);
while(q --)work();
}
return 0;
}
O(n^2)的线性dp 不要想太多。。(我都想到DAG去了。。。)
代码比较简单,可以直接看懂。
#include <bits/stdc++.h>
using namespace std;
vector <int> ps [5010], cs[5010];
int s[5010] , e[5010];
bool vis[5010];
long long dp [5010];
int main()
{
int n;
scanf("%d",&n);
vector <int> a (n);
for (int i = 0; i < n; i++) {
scanf("%d",&a[i]);
if (!vis[a[i]]) s[a[i]] = i;
vis[a[i]] = 1;
e[a[i]] = i;
}
for (int i = 0; i < n; i++) vis[a[i]] = 0;
for (int i = 0; i < n; i++) {
int mx = i, rx = 0, lj = i;
for (int j = i; j < n; j++) {
if (s[a[j]] < i) break;
lj = j;
mx = max(mx,e[a[j]]);
if (!vis[a[j]]) rx ^= a[j];
vis[a[j]] = 1;
if (mx <= j) {
ps[i].push_back(j);
cs[i].push_back(rx);
}
}
for (int k = lj; k >= i; k--) vis[a[k]] = 0;
}
for (int i = n-1; i >= 0; i--) {
dp[i] = dp[i+1];
for (int j = 0; j < ps[i].size(); j++) {
dp[i] = max(dp[i],dp[ps[i][j]+1]+cs[i][j]);
}
}
printf("%I64d\n",dp[0]);
}
E. Vladik and Entertaining Flags
原文:http://www.cnblogs.com/heisenberg-/p/6914443.html