A Crossmarket
思维
矩阵走路径，发现走Z字型怎么走都是一样的耗费，所以直接O(1)算出来就好
/*
* |~~~~~~~|
* | |
* | |
* | |
* | |
* | |
* |~.\\\_\~~~~~~~~~~~~~~xx~~~ ~~~~~~~~~~~~~~~~~~~~~/_//;~|
* | \ o \_ ,XXXXX), _..-~ o / |
* | ~~\ ~-. XXXXX`)))), _.--~~ .-~~~ |
* ~~~~~~~`\ ~\~~~XXX' _/ ';)) |~~~~~~..-~ _.-~ ~~~~~~~
* `\ ~~--`_\~\, ;;;\)__.---.~~~ _.-~
* ~-. `:;;/;; \ _..-~~
* ~-._ `'' /-~-~
* `\ / /
* | , | |
* | ' / |
* \/; |
* ;; |
* `; . |
* |~~~-----.....|
* | \ \
* | /\~~--...__ |
* (| `\ __-\|
* || \_ /~ |
* |) \~-' |
* | | \ '
* | | \ :
* \ | | |
* | ) ( )
* \ /; /\ |
* | |/ |
* | | |
* \ .' ||
* | | | |
* ( | | |
* | \ \ |
* || o `.)|
* |`\\) |
* | |
* | |
*/
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<map>
// #pragma GCC optimize(3)
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define scan(x) scanf("%lld",&x)
#define int long long
#define lowbit(x) x&(-x) //二进制最低位所代表的数
#define PI 3.1415926535
typedef pair<int,int> PII;
int gcd(int a,int b){
return b>0 ? gcd(b,a%b):a;
}
int exgcd(int a,int b,int &x,int &y)
{
if(!b)
{
x = 1,y = 0;
return a;
}
int d = exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
const int N = 1e6+10;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
void init()
{
cin.tie(0),cout.tie(0);
ios::sync_with_stdio(false);
}
void solve()
{
int n,m;
cin>>n>>m;
if(n<m)swap(n,m);
if(n==1&&m==1)
{
cout<<0<<endl;
return;
}
int t = n+(m-2)+2+(m-2);
cout<<t<<endl;
}
signed main()
{
init();
int t;
cin>>t;
while(t--)solve();
}
View Code
B Beautiful a Array
构造
放k*b就满足了“漂亮”性质，然后还要每个都加上min(s,k-1)来满足总数和的性质
/*
* |~~~~~~~|
* | |
* | |
* | |
* | |
* | |
* |~.\\\_\~~~~~~~~~~~~~~xx~~~ ~~~~~~~~~~~~~~~~~~~~~/_//;~|
* | \ o \_ ,XXXXX), _..-~ o / |
* | ~~\ ~-. XXXXX`)))), _.--~~ .-~~~ |
* ~~~~~~~`\ ~\~~~XXX' _/ ';)) |~~~~~~..-~ _.-~ ~~~~~~~
* `\ ~~--`_\~\, ;;;\)__.---.~~~ _.-~
* ~-. `:;;/;; \ _..-~~
* ~-._ `'' /-~-~
* `\ / /
* | , | |
* | ' / |
* \/; |
* ;; |
* `; . |
* |~~~-----.....|
* | \ \
* | /\~~--...__ |
* (| `\ __-\|
* || \_ /~ |
* |) \~-' |
* | | \ '
* | | \ :
* \ | | |
* | ) ( )
* \ /; /\ |
* | |/ |
* | | |
* \ .' ||
* | | | |
* ( | | |
* | \ \ |
* || o `.)|
* |`\\) |
* | |
* | |
*/
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<map>
// #pragma GCC optimize(3)
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define scan(x) scanf("%lld",&x)
#define int long long
#define lowbit(x) x&(-x) //二进制最低位所代表的数
#define PI 3.1415926535
typedef pair<int,int> PII;
int gcd(int a,int b){
return b>0 ? gcd(b,a%b):a;
}
int exgcd(int a,int b,int &x,int &y)
{
if(!b)
{
x = 1,y = 0;
return a;
}
int d = exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
const int N = 1e6+10;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
void init()
{
cin.tie(0),cout.tie(0);
ios::sync_with_stdio(false);
}
void solve()
{
int n,k,b,s;
cin>>n>>k>>b>>s;
vector<int> a(n);
a[0]=k*b;
s-=k*b;
if(s<0)
{
cout<<-1<<endl;
return;
}
for(int i=0;i<n;i++)
{
int t = min(s,k-1);
a[i]+=t;
s-=t;
}
if(s>0)
{
cout<<-1<<endl;
}
else
{
for(int i=0;i<n;i++)
{
cout<<a[i]<<' ';
}
cout<<endl;
}
}
signed main()
{
init();
int t;
cin>>t;
while(t--)solve();
}
View Code
C Monoblock
数学规律
这种查询题就是先找出最终结果然后进行减法操作，这样才不会超时。
所以这道题的触发是换之后是否相等，相等的话那么会进行减法，减去有这两个元素的区间，不相等的话加上。
子区间( i )*( n - i )
/*
* |~~~~~~~|
* | |
* | |
* | |
* | |
* | |
* |~.\\\_\~~~~~~~~~~~~~~xx~~~ ~~~~~~~~~~~~~~~~~~~~~/_//;~|
* | \ o \_ ,XXXXX), _..-~ o / |
* | ~~\ ~-. XXXXX`)))), _.--~~ .-~~~ |
* ~~~~~~~`\ ~\~~~XXX' _/ ';)) |~~~~~~..-~ _.-~ ~~~~~~~
* `\ ~~--`_\~\, ;;;\)__.---.~~~ _.-~
* ~-. `:;;/;; \ _..-~~
* ~-._ `'' /-~-~
* `\ / /
* | , | |
* | ' / |
* \/; |
* ;; |
* `; . |
* |~~~-----.....|
* | \ \
* | /\~~--...__ |
* (| `\ __-\|
* || \_ /~ |
* |) \~-' |
* | | \ '
* | | \ :
* \ | | |
* | ) ( )
* \ /; /\ |
* | |/ |
* | | |
* \ .' ||
* | | | |
* ( | | |
* | \ \ |
* || o `.)|
* |`\\) |
* | |
* | |
*/
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<stack>
#include<unordered_map>
#include<unordered_set>
#include<set>
#include<map>
// #pragma GCC optimize(3)
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define scan(x) scanf("%lld",&x)
#define int long long
#define lowbit(x) x&(-x) //二进制最低位所代表的数
#define PI 3.1415926535
typedef pair<int,int> PII;
int gcd(int a,int b){
return b>0 ? gcd(b,a%b):a;
}
int exgcd(int a,int b,int &x,int &y)
{
if(!b)
{
x = 1,y = 0;
return a;
}
int d = exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
const int N = 1e6+10;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
void init()
{
cin.tie(0),cout.tie(0);
ios::sync_with_stdio(false);
}
int a[N];
void solve()
{
int n,m;
cin>>n>>m;
for(int i=1;i<=n;i++)
{
cin>>a[i];
}
int cnt = 0;
int sum = 0;
for(int i=1;i<=n;i++)
{
if(a[i]!=a[i-1])cnt+=i;
else cnt++;
sum+=cnt;
}
while(m--)
{
int i,x;
cin>>i>>x;
if(a[i-1]!=a[i]&&a[i-1]==x)
{
sum-=(i-1)*(n-i+1);
}
if(a[i-1]==a[i]&&a[i-1]!=x)
{
sum+=(i-1)*(n-i+1);
}
if(a[i+1]!=a[i]&&a[i+1]==x)
{
sum-=(i)*(n-i);
}
if(a[i+1]==a[i]&&a[i+1]!=x)
{
sum+=(i)*(n-i);
}
a[i]=x;
cout<<sum<<endl;
}
}
signed main()
{
init();
int t=1;
while(t--)solve();
}
View Code
标签：const,int,void,CF,_.-,return,include,div2,816 From： https://www.cnblogs.com/yeonnyuui/p/16609889.html