Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2084 Accepted Submission(s): 1348
Problem Description
Connecting the display screen and signal sources which produce different color signals by cables, then the display screen can show the color of the signal source.Notice that every signal source can only send signals to one display screen each time.
Now you have K different colors.
Now you have K different colors.
Input
Multiple cases (no more than K.
Output
Output the minimum number of cables N.
Sample Input
3 2
20 15
Sample Output
4
90
Hint
就是这两个数字组合下嘛，很容易直接观察出结果的
#include <bits/stdc++.h> using namespace std; typedef long long ll; typedef unsigned long long ull; const int N = 1e6+5; const int INF=0x3f3f3f3f; int main() { ll n,k; while(cin>>n>>k) { cout<<k*(n-k+1)<<endl; } return 0; }
array array array
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2917 Accepted Submission(s): 1170
Problem Description
One day, Kaitou Kiddo had stolen a priceless diamond ring. But detective Conan blocked Kiddo's path to escape from the museum. But Kiddo didn't want to give it back. So, Kiddo asked Conan a question. If Conan could give a right answer, Kiddo would return the ring to the museum.
Kiddo: "I have an array
Kiddo: "I have an array
Input
The first line contains an integer T indicating the total number of test cases. Each test case starts with two integers 5
Output
For each test case, please output "A is a magic array." if it is a magic array. Otherwise, output "A is not a magic array." (without quotes).
Sample Input
3
4 1
1 4 3 7
5 2
4 1 3 1 2
6 1
1 4 3 5 4 6
Sample Output
A is a magic array.
A is a magic array.
A is not a magic array.
最长递增子序列
#include<iostream> #include<algorithm> #include<stdio.h> using namespace std; const int N=1e6+10; const int INF=0x3f3f3f3f; int a[N],g[N],f[N],b[N],c[N],n,h[N]; int main() { int t,n,last,l,i,m,x; cin>>t; while(t--) { cin>>n>>m; for(int i=0; i<n; i++) cin>>a[i],h[n-i-1]=a[i]; fill(g,g+n,INF); b[0]=1; for(int i=0; i<n; i++) { int j=lower_bound(g, g+n,a[i])-g; g[j]=a[i]; b[i]=j; } l=lower_bound(g, g+n,INF)-g-1; last=INF; for(i=n-1;i>=0;i--) { if(l==-1)break; if(b[i]==l&&a[i]<last) { last=a[i]; c[l]=last; l--; } } l=lower_bound(g, g+n,INF)-g; l=n-l;x=l; fill(g,g+n,INF); b[0]=1; for(int i=0; i<n; i++) { int j=lower_bound(g, g+n,h[i])-g; g[j]=h[i]; b[i]=j; } l=lower_bound(g, g+n,INF)-g-1; last=INF; for(i=n-1;i>=0;i--) { if(l==-1)break; if(b[i]==l&&a[i]<last) { last=a[i]; c[l]=last; l--; } } l=lower_bound(g, g+n,INF)-g; if(x>n-l) x=n-l; if(x<=m) puts("A is a magic array."); else puts("A is not a magic array."); //printf("%d\n",l); } return 0; }
number number number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2021 Accepted Submission(s): 1068
Problem Description
We define a sequence d number.
The answer may be too large. Please print the answer modulo 998244353.
The answer may be too large. Please print the answer modulo 998244353.
Input
There are about 500 test cases, end up with EOF.
Each test case includes an integer 9)
Each test case includes an integer 9)
Output
For each case, output the minimal d number mod 998244353.
Sample Input
1
Sample Output
4
找规律+矩阵快速幂
#include<cstdio> #include<algorithm> #include<cstring> #include<iostream> using namespace std; const long long M =998244353; struct Matrix { long long a[2][2]; Matrix() { memset(a, 0, sizeof(a)); } Matrix operator * (const Matrix y) { Matrix ans; for(long long i = 0; i <= 1; i++) for(long long j = 0; j <= 1; j++) for(long long k = 0; k <= 1; k++) ans.a[i][j] += a[i][k]*y.a[k][j]; for(long long i = 0; i <= 1; i++) for(long long j = 0; j <= 1; j++) ans.a[i][j] %= M; return ans; } void operator = (const Matrix b) { for(long long i = 0; i <= 1; i++) for(long long j = 0; j <= 1; j++) a[i][j] = b.a[i][j]; } }; long long solve(long long x) { Matrix ans, trs; ans.a[0][0] = ans.a[1][1] = 1; trs.a[0][0] = trs.a[1][0] = trs.a[0][1] = 1; while(x) { if(x&1) ans = ans*trs; trs = trs*trs; x >>= 1; } return ans.a[0][0]; } int main() { long long n; while(~scanf("%lld", &n)) { cout <<(solve(2*n+2)-1+M)%M << endl; } return 0; }
transaction transaction transaction
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 132768/132768 K (Java/Others)
Total Submission(s): 2245 Accepted Submission(s): 682
Problem Description
Kelukin is a businessman. Every day, he travels around cities to do some business. On August 17th, in memory of a great man, citizens will read a book named "the Man Who Changed China". Of course, Kelukin wouldn't miss this chance to make money, but he doesn't have this book. So he has to choose two city to buy and sell.
As we know, the price of this book was different in each city. It is n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.
As we know, the price of this book was different in each city. It is n cities. Kelukin can choose any city to start his travel. He want to know the maximum money he can get.
Input
The first line contains an integer ).
Output
For each test case, output a single number in a line: the maximum money he can get.
Sample Input
1
4
10 40 15 30
1 2 30
1 3 2
3 4 10
Sample Output
8
dfs直接维护下两个值的大小就好，别来理解错题意
#include<bits/stdc++.h> using namespace std; const int N=1e5+5; vector<pair<int,int> >G[N]; int sell[N],buy[N],ans; int vis[N]; void dfs(int x) { for(int i=0;i<(int)G[x].size();i++) { int v=G[x][i].first; int w=G[x][i].second; if(vis[v])continue; vis[v]=1; dfs(v); buy[x]=min(buy[x],buy[v]+w); sell[x]=max(sell[x],sell[v]-w); } ans=max(ans,buy[x]-sell[x]); ans=max(ans,sell[x]-buy[x]); } int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); memset(vis,0,sizeof(vis)); for(int i=1;i<=n;i++) G[i].clear(); for(int i=1;i<=n;i++) { scanf("%d",&sell[i]); buy[i]=sell[i]; } for(int i=1;i<n;i++) { int u,v,w; scanf("%d%d%d",&u,&v,&w); G[u].push_back({v,w}); G[v].push_back({u,w}); } ans=0; vis[1]=1; dfs(1); printf("%d\n",ans); } return 0; }
card card card
Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2684 Accepted Submission(s): 755
Problem Description
As a fan of Doudizhu, WYJ likes collecting playing cards very much.
One day, MJF takes a stack of cards and talks to him: let's play a game and if you win, you can get all these cards. MJF randomly assigns these cards into
One day, MJF takes a stack of cards and talks to him: let's play a game and if you win, you can get all these cards. MJF randomly assigns these cards into
Input
There are about i.
Output
For each test case, print only an integer, denoting the number of piles WYJ needs to move before the game starts. If there are multiple solutions, print the smallest one.
Sample Input
5
4 6 2 8 4
1 5 7 9 2
Sample Output
4
Hint
For the sample input: + If WYJ doesn't move the cards pile, when the game starts the state of cards is: 4 6 2 8 4 1 5 7 9 2 WYJ can take the first three piles of cards, and during the process, the number of face-up cards is 4-1+6-5+2-7. Then he can't pay the the "penalty value" of the third pile, the game ends. WYJ will get 12 cards. + If WYJ move the first four piles of cards to the end, when the game starts the state of cards is: 4 4 6 2 8 2 1 5 7 9 WYJ can take all the five piles of cards, and during the process, the number of face-up cards is 4-2+4-1+6-5+2-7+8-9. Then he takes all cards, the game ends. WYJ will get 24 cards. It can be improved that the answer is 4. **huge input, please use fastIO.**
直接模拟下，但是有个细节就是全部都可以取到
#include<stdio.h> #include<algorithm> #include<string.h> using namespace std; int a[1000005],b[1000005]; int main() { int n,x,i,s,p,y,q,r; while(~scanf("%d",&n)) { for(i=0;i<n;i++) scanf("%d",&a[i]),b[i]=a[i]; for(i=0;i<n;i++) { scanf("%d",&x); a[i]-=x; } s=0;p=i;y=0;q=0;r=0; for(i=0;i<n;i++) { s+=a[i]; if(s<0) { if(q<y) { q=y,r=p; } y=0; s=0; p=i+1; } y+=b[i]; } if(p==n) { printf("%d\n",r); continue; } for(i=0;i<p;i++) { s+=a[i]; if(s<0) { if(q<y) { q=y,r=p; } break; } y+=b[i]; } if(q<y) { q=y,r=p; } printf("%d\n",r); } }