Problem A Oshino Shinobu loves Mr.Donut
https://ac.nowcoder.com/acm/contest/908/A
题意:
题解:
C++版本一
Problem B Kuangyeye's Resistance
https://ac.nowcoder.com/acm/contest/908/B
题意:计算n级网络的等效电阻。
题解:
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
ll t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
ll POW(ll a,ll b=p-2,ll c=p){
ll res=1;
ll base=a%c;
while(b){
if(b&1)res=(res*base)%p;
base=(base*base)%p;
b>>=1;
}
return res;
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%lld%lld%lld",&n,&r,&p);
ll R=r;
for(int i=2;i<=n;i++){
R=(((R+2*r)%p*r%p)*((POW(R+3*r,p-2,p))%p))%p;
}
cout<<R<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem C One Piece
https://ac.nowcoder.com/acm/contest/908/C
题意:二进制表示中找到1的数。
题解:__builtin_popcount
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
scanf("%d",&t);
while(t--){
scanf("%d",&n);
printf("%d\n",__builtin_popcount(n));
}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem D Kth height
https://ac.nowcoder.com/acm/contest/908/D
题意:两只球队,都是n个人且都身高递增,m次操作,将p队的第q个人的身高变为t,并询问两只球队总集合中第k矮的那个人的身高,保证修改之后依然递增
题解:二分
C++版本一
题解:二分
1、二分出前k个人,在第一队里面有多少个;
2、注意多组数据;
3、二分范围[0,k];
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v,q;
int ans,cnt,flag,temp,sum;
int a[3][N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
while(~scanf("%d",&n)){
for(int i=1;i<=n;i++)scanf("%d",&a[1][i]);
for(int i=1;i<=n;i++)scanf("%d",&a[2][i]);
scanf("%d",&m);
for(int i=1;i<=m;i++){
scanf("%d%d%d%d",&p,&q,&t,&k);
a[p][q]=t;
l=0;r=k;
while(l<=r){
int mid=(l+r)>>1;
int temp=(upper_bound(a[2]+1,a[2]+n+1,a[1][mid])-a[2])-1;
if(mid+temp<=k){
ans=mid;
l=mid+1;
}else{
r=mid-1;
}
}
cout<<max(a[1][ans],a[2][k-ans])<<endl;
}
}
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem E Just calculat
https://ac.nowcoder.com/acm/contest/908/E
题意:
题解:
C++版本一
Problem F Cards with Numbers
https://ac.nowcoder.com/acm/contest/908/F
题意:
两个操作:
- 0 x( 表示在包装盒中放入带有数字x的卡片。)
- 1 x( 表示查询在包装盒中是否有号码为x的卡片。)
题解:
1、字典树
2、离散化+离线处理
C++版本一
题解:字典树
注意:数据原因可能还没最糟糕的情况所以卡过
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=2500000+10;
const int M=1000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v,op;
int ans,cnt,flag,temp;
int a[N][10];
int en[N];
char x[20];
void insert(char *s){
int len=strlen(s),p=0;
for(int i=0;i<len;i++){
int ch=s[i]-'0';
if(a[p][ch]==-1)a[p][ch]=++cnt;
p=a[p][ch];
}
en[p]=1;
}
bool search(char* s){
int len=strlen(s),p=0;
for(int i=0;i<len;i++){
int ch=s[i]-'0';
if(a[p][ch]==-1)
return 0;
p=a[p][ch];
}
return en[p];
}
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
memset(a,-1,sizeof(a));
for(int i=1;i<=n;i++){
scanf("%d%s",&op,x);
if(!op){
insert(x);
}else{
cout<<(search(x)?"yes":"no")<<endl;
}
}
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
C++版本二
题解: 离散化+离线处理
Problem G Longest Palindrome Substring
https://ac.nowcoder.com/acm/contest/908/G
题意:最长的回文子串的长度是否大于1
题解:枚举
C++版本一
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str[N];
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d",&n);
cin>>str;
for(int i=1;i<n-1;i++){
if(str[i-1]==str[i+1]||str[i-1]==str[i]||str[i]==str[i+1]){
ans=1;
break;
}
}
cout<<(ans||(n==2&&str[0]==str[1])?"YES":"NO")<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem H Longest Common Palindrome Substring
https://ac.nowcoder.com/acm/contest/908/H
题意:
题解:
C++版本一
Problem I Algorithm Choosing Mushrooms
https://ac.nowcoder.com/acm/contest/908/I
题意:给你n个篮子以及篮子里有的蘑菇,你可以取走一段连续的篮子,但这些篮子里的蘑菇总和必须是m的倍数,问你最多拿几只篮子?最多拿多少蘑菇?
题解:DP
C++版本一
题解:DP
1、记录每个余数第一次出现的位置;
2、余数再次出现,则以当前数为结尾的最长区间【第一次位置,当前位置】;
/*
*@Author: STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=200000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
ll ans,cnt,flag,temp,sum;
ll a[N];
ll b[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
#endif
//ios::sync_with_stdio(false);
//cin.tie(0);
//cout.tie(0);
//scanf("%d",&t);
//while(t--){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
a[i]+=a[i-1];
temp=a[i]%m;
if(b[temp]==0)
b[temp]=i;
if((a[i]-a[b[temp]])%m==0){
ans=max(ans,i-b[temp]);
sum=max(sum,a[i]-a[b[temp]]);
}
}
cout<<sum<<" "<<ans<<endl;
//}
#ifdef DEBUG
printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
//cout << "Hello world!" << endl;
return 0;
}
Problem J Simple Problem
https://ac.nowcoder.com/acm/contest/908/J
题意:
题解:
C++版本一
Problem K Water Problem
https://ac.nowcoder.com/acm/contest/908/K
题意:
题解:
C++版本一