The Preliminary Contest for ICPC China Nanchang 南昌网络赛 A H I K M J题
我补过 J题拉 orz 交了3次 重写了一次orz
真是太快乐(自闭)拉
连接 https://blog.csdn.net/qq_40831340/article/details/90739880
A PERFECT NUMBER PROBLEM
Write a program to output the first 55 perfect numbers. A perfect number is defined to be a positive integer where the sum of its positive integer divisors excluding the number itself equals the number.
For example: 1+ 2 + 3 = 61+2+3=6, and 66 is the first perfect number.
There is no input for this problem.
百度完美数可得。。。。
int main(){
cout<<"6"<<endl;
cout<<"28"<<endl;
cout<<"496"<<endl;
cout<<"8128"<<endl;
cout<<"33550336"<<endl;
return 0;
}
M Subsequence
给母串 和一些子串 问 字串是否在母串出现过 (可以不连续) 2018百度之星网络赛第一题 2019 牛客小白月赛12 最后一题 贴过来就过 。。。
其实就是开了个26*maxn 的数组 倒着扫 存每层下一个字母出现位置 就可以 o(m) 的确定一个串存不存在了
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+5;
char mp[maxn];
char sp[maxn];
int pos[30][maxn];
int main() {
scanf("%s",mp);
int len=strlen(mp);
for(int i=len-1; i>=0; i--) {
for(int j=27; j>=1; j--) {
pos[j][i]=pos[j][i+1];
}
pos[mp[i]-'a'+1][i]=i+1;
}
int n;
scanf("%d",&n);
for(int i=1; i<=n; i++) {
scanf("%s",sp);
len=strlen(sp);
int j=0;
int nxt=0;
while(j<len) {
nxt=pos[sp[j]-'a'+1][nxt];
if(nxt==0) {
cout<<"NO\n";
break;
}
if(j==len-1) {
cout<<"YES\n";
break;
}
j++;
}
}
return 0;
}
H Coloring Game
David has a white board with 2 \times N2×N grids.He decides to paint some grids black with his brush.He always starts at the top left corner and ends at the bottom right corner, where grids should be black ultimately.
Each time he can move his brush up(↑), down(↓), left(←), right(→), left up(↖), left down(↙), right up(↗), right down (↘) to the next grid.
For a grid visited before,the color is still black. Otherwise it changes from white to black.
David wants you to compute the number of different color schemes for a given board. Two color schemes are considered different if and only if the color of at least one corresponding position is different.
给你一个2x n 个网格 这个人左上走到右下 走过的格子涂黑 它可以走周围8个方向
因为它最后是到右下的 我们可以认为 这个2x n 除了第一层(左上一定黑) 和 最后一层（右下一定黑）
其他层 只有3个状态 10 01 11 所以 是3^（n-2）个状态 + 第一层和最后一次的4个状态
ans = 4*3^（n-2）;
ps 1列 特判；
#include<bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
//#define int long long
using namespace std;
typedef long long ll;
const long long INF = 0x3f3f3f3f3f3f3f3f ;
const int maxn = 5e5 + 5 ;
const long long mod=1000000007;
long long int quick_mod(long long int a,long long int b) {
long long int ans=1;
a%=mod;
while(b) {
if(b&1) {
ans=ans*a%mod;
b--;
}
b>>=1;
a=a*a%mod;
}
return ans;
}
int main() {
ll n;
cin>>n;
if(n==1) cout<<1<<endl;
else cout<<4*quick_mod(3,n-2)%mod<<endl;
return 0;
}
I Max answer
Alice has a magic array. She suggests that the value of a interval is equal to the sum of the values in the interval, multiplied by the smallest value in the interval.
Now she is planning to find the max value of the intervals in her array. Can you help her?
给你一个长度n的序列 选取一个数字 和(包含它连续的)区间(而且它必须是这个区间的最小值) 使a[i]*(这一区间的和) 最大
首先单调栈 处理每个a[ i ] 管理区间(没有比它小的最大连续区间)
然后 对 a[ i ] 分2种情况 正负来考虑
- 大于 0 的情况 显然是 它 管理的区间和*a[ i ] 因为a[i]>0 同时它肯定是这区间最低点 该区间越长和越大
- 小于 0 的情况 不妨我们维护一个 前缀和 既然它是负数 我们 当前i 到R[ i ]前缀和最好越小越好 而L[ i ] 到 i得和越大越好
#include<bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
//#define int long long
using namespace std;
typedef long long ll;
const long long INF = 0x3f3f3f3f3f3f3f3f ;
const int maxn = 5e5 + 5 ;
int n;
ll a[maxn],sum[maxn];
int L[maxn],R[maxn];
struct node {
ll minsum,maxsum;
} tr[maxn<<4];
void pushup(int rt) {
tr[rt].minsum=min(tr[rt<<1].minsum,tr[rt<<1|1].minsum);
tr[rt].maxsum=max(tr[rt<<1].maxsum,tr[rt<<1|1].maxsum);
}
void build(int l,int r,int rt) {
if(l==r) {
tr[rt]={sum[l],sum[l]};
return ;
}
int mid=(l+r)>>1;
build(l,mid,rt<<1);
build(mid+1,r,rt<<1|1);
pushup(rt);
}
ll maxquery(int L,int R,int l,int r,int rt){
if(L<=l&&r<=R){
return tr[rt].maxsum;
}
int mid=(l+r)>>1;
ll res=-INF;
if(L<=mid) res=max(res,maxquery(L,R,l,mid,rt<<1));
if(R>mid) res=max(res,maxquery(L,R,mid+1,r,rt<<1|1));
return res;
}
ll minquery(int L,int R,int l,int r,int rt){
if(L<=l&&r<=R){
return tr[rt].minsum;
}
int mid=(l+r)>>1;
ll res=INF;
if(L<=mid) res=min(res,minquery(L,R,l,mid,rt<<1));
if(R>mid) res=min(res,minquery(L,R,mid+1,r,rt<<1|1));
return res;
}
void solLR() {
stack<int> s;
for(int i=1; i<=n; i++) {
while(!s.empty()&&a[s.top()]>=a[i]) s.pop();
L[i]=(s.empty())?0:s.top();
s.push(i);
}
while(!s.empty()) s.pop();
for(int i=n; i>=1; i--) {
while(!s.empty()&&a[s.top()]>=a[i]) s.pop();
R[i]=(s.empty())?(n+1):s.top();
s.push(i);
}
}
signed main() {
scanf("%lld",&n);
for(int i=1; i<=n; i++) scanf("%lld",&a[i]),sum[i]=sum[i-1]+a[i];
solLR();
build(1,n,1);
ll ans=-INF;
ll rmin,lmax;
for(int i=1; i<=n; i++) {
if(a[i]<0) {
rmin=minquery(i,R[i]-1,1,n,1);
lmax=maxquery(L[i]+1,i,1,n,1);
ans=max(ans,a[i]*(rmin-lmax));
} else {
ans=max(ans,a[i]*(sum[R[i]-1]-sum[L[i]]));
}
}
printf("%lld\n",ans);
return 0;
}
K MORE XOR
Given a sequence of n numbers and three functions.
Define a function f(l,r)f(l,r) which returns \oplus a[x]⊕a[x] (l \le x \le rl≤x≤r). The \oplus⊕ represents exclusive OR.
Define a function g(l,r)g(l,r) which returns \oplus f(x,y)(l \le x \le y \le r)⊕f(x,y)(l≤x≤y≤r).
Define a function w(l,r)w(l,r) which returns \oplus g(x,y)(l \le x \le y \le r)⊕g(x,y)(l≤x≤y≤r).
You are also given a number of xor-queries. A xor-query is a pair (i, ji,j) (1 \le i \le j \le n1≤i≤j≤n). For each xor-query (i, j)(i,j), you have to answer the result of function w(l,r)w(l,r).
让你求异或得前缀和的前缀和的前缀和 orz 规律题 异或 出现2次的数字 显然是0 这里搞贡献度就好
暴力直接for超时了 这里选择维护一个前缀异或和的数组
#include<bits/stdc++.h>
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f ;
const int maxn = 1e5 + 5 ;
int a[maxn];
int main() {
int t,n,l,r,res,len,q;
scanf("%d",&t);
while(t--) {
scanf("%d",&n);
for(int i=1; i<=n; i++) scanf("%d",&a[i]);
for(int i=4; i<=n; i++) a[i]^=a[i-4];
scanf("%d",&q);
while(q--) {
scanf("%d %d",&l,&r);
len = r - l + 1 ;
if(len%4==0) printf("0\n");
else if(len % 4 == 1) {
if(l>=4) res=a[r]^a[l-4];
else res=a[r];
printf("%d\n",res);
} else if(len % 4 == 2) {
if(l>=4) res=a[r]^a[l-4];
else res=a[r];
l++,r--;
if(l>=4) res=res^a[r]^a[l-4];
else res=res^a[r];
printf("%d\n",res);
} else if(len % 4 == 3) {
l++,r--;
if(l>=4) res=a[r]^a[l-4];
else res=a[r];
printf("%d\n",res);
}
}
}
return 0;
}
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