Now ,there are some rectangles. The area of these rectangles is 1* x or 2 * x ,and now you need find a big enough rectangle( 2 * m) so that you can put all rectangles into it(these rectangles can't rotate). please calculate the minimum m satisfy the condition.

There are some tests ,the first line give you the test number. 
Each test will give you a number n (1<=n<=100)show the rectangles number .The following n rows , each row will give you tow number a and b. (a = 1 or 2 , 1<=b<=100).

Each test you will output the minimum number m to fill all these rectangles.

#include<stdio.h>
const int N = 10005;
int main()
{int dp[N],ans,t,a[N],b[N],n;scanf("%d",&t);while(t--){scanf("%d",&n);ans=0;for(int i=0;i<=N;i++)dp[i]=0;int V=0;for(int i=1;i<=n;i++){scanf("%d%d",&a[i],&b[i]);if(a[i]==1)V+=b[i];}for(int i=1;i<=n;i++)if(a[i]==2)ans+=b[i];else{for(int v=V/2;v>=b[i];v--)if(dp[v]<dp[v-b[i]]+b[i])dp[v]=dp[v-b[i]]+b[i];}int aa=V-dp[V/2];if(aa<dp[V/2])aa=dp[V/2];ans+=aa;printf("%d\n",ans);}return 0;
}/*
2
3
1 2
2 2
2 3
3
1 2
1 2
1 3
*/

You need to design road from (0, 0) to (x, y) in plane with the lowest cost. Unfortunately, there are N Rivers between (0, 0) and (x, y).It costs c1 Yuan RMB per meter to build road, and it costs c2 Yuan RMB per meter to build a bridge. All rivers are parallel to the Y axis with infinite length.

There are several test cases.
Each test case contains 5 positive integers N,x,y,c1,c2 in the first line.(N ≤ 1000,1 ≤ x,y≤ 100,000,1 ≤ c1,c2 ≤ 1000).
The following N lines, each line contains 2 positive integer xi, wi ( 1 ≤ i ≤ N ,1 ≤ xi ≤x, xi-1+wi-1 < xi , xN+wN ≤ x),indicate the i-th river(left bank) locate xi with wi width.
The input will finish with the end of file.

For each the case, your program will output the least cost P on separate line, the P will be to two decimal places .

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;int n;
double x,y,c1,c2,sum,ans;double dis(double x1,double y1,double x2,double y2)
{return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}void fen_3(double l,double r)
{double mid1 = l+(r-l)/3;double mid2 = l+(r-l)*2/3;double s1 = dis(0,0,sum,mid1)*c2+dis(sum,mid1,x,y)*c1;double s2 = dis(0,0,sum,mid2)*c2+dis(sum,mid2,x,y)*c1;if(fabs(s1-s2)<1e-6){ans = s2;return;}if(s1>s2)fen_3(mid1,r);elsefen_3(l,mid2);
}int main()
{int i,j,k;while(~scanf("%d%lf%lf%lf%lf",&n,&x,&y,&c1,&c2)){sum = 0;for(i = 0;i<n;i++){double tx,ty;scanf("%lf%lf",&tx,&ty);sum+=ty;}fen_3(0,y);printf("%.2f\n",ans);}return 0;
}/**************************************************************Problem: 1548User: aking2015Language: C++Result: AcceptedTime:84 msMemory:976 kb
****************************************************************/

Navigation is a field of study that focuses on the process of monitoring and controlling the movement of a craft or vehicle from one place to another. The field of navigation includes four general categories: land navigation, marine navigation, aeronautic navigation, and space navigation. (From Wikipedia)
Recently, slowalker encountered a problem in the project development of Navigition APP. In order to provide users with accurate navigation service , the Navigation APP should re-initiate geographic location after the user walked DIST meters. Well, here comes the problem. Given the Road Information which could be abstracted as N segments in two-dimensional space and the distance DIST, your task is to calculate all the postion where the Navigition APP should re-initiate geographic location.

The input file contains multiple test cases.
For each test case,the first line contains an interger N and a real number DIST. The following N+1 lines describe the road information.
You should proceed to the end of file.

Hint:
1 <= N <= 1000
0 < DIST < 10,000

For each the case, your program will output all the positions where the Navigition APP should re-initiate geographic location. Output “No Such Points.” if there is no such postion.

Welcome,this is the 2015 3th Multiple Universities Programming Contest ,Changsha ,Hunan Province. In order to let you feel fun, ACgege will give you a simple problem. But is that true? OK, let’s enjoy it.
There are three strings A , B and C. The length of the string A is 2*N, and the length of the string B and C is same to A. You can take N characters from A and take N characters from B. Can you set them to C ?

There are several test cases.
Each test case contains three lines A,B,C. They only contain upper case letter.
0<N<100000
The input will finish with the end of file.

For each the case, if you can get C, please print “YES”. If you cann’t get C, please print “NO”.

#include<stdio.h>
#include<string.h>
const int N = 200005;char A[N],B[N],C[N];int main()
{int a[30],b[30],c[30];int mint,maxt,flag,len;while(scanf("%s%s%s",A,B,C)>0){len=strlen(A);for(int i=0;i<=26;i++)a[i]=b[i]=c[i]=0;for(int i=0;i<len;i++){int j;j=A[i]-'A'; a[j]++;j=B[i]-'A'; b[j]++;j=C[i]-'A'; c[j]++;}flag=0;for(int i=0;i<26;i++){if(a[i]>len/2)a[i]=len/2;if(b[i]>len/2)b[i]=len/2;if(a[i]+b[i]<c[i]){flag=1; break;}}if(flag){printf("NO\n");continue;}mint=maxt=0;for(int i=0;i<26;i++){if(c[i]>b[i])mint+=c[i]-b[i];if(a[i]>=c[i])maxt+=c[i];elsemaxt+=a[i];}//       printf("%d   %d\n",mint,maxt);if(len/2>=mint&&len/2<=maxt)printf("YES\n");elseprintf("NO\n");}
}/**************************************************************Problem: 1550User: aking2015Language: C++Result: AcceptedTime:24 msMemory:1548 kb
****************************************************************/

Give you a numeric sequence. If you can demolish arbitrary amount of numbers, what is the length of the longest increasing sequence, which is made up of consecutive numbers? It sounds like Longest Increasing Subsequence at first sight. So, there is another limitation: the numbers you deleted must be consecutive.

There are several test cases.
For each test case, the first line of input contains the length of sequence N(1≤N≤10^4). The second line contains the elements of sequence——N positive integers not larger than 10^4.

For each the case, output one integer per line, denoting the length of the longest increasing sequence of consecutive numbers, which is achievable by demolishing some(may be zero) consecutive numbers.

转载请注明出处：寻找&星空の孩子

#include<stdio.h>
#include<algorithm>
using namespace std;
const int N = 10005;
struct qj
{int l,r;
}kuai[N];
int b[N],tn,n,a[N],k[N],tree[N*4];
void builde(int l,int r,int e)
{int m=(l+r)/2;tree[e]=0;if(l==r)return ;builde(l,m,e*2);builde(m+1,r,e*2+1);
}
void settree(int l,int r,int e,int id,int ans)
{int m=(l+r)/2;if(tree[e]<ans)tree[e]=ans;if(l==r){return ;}if(id<=m)settree(l,m,e*2,id,ans);elsesettree(m+1,r,e*2+1,id,ans);
}
int findans(int l,int r,int e,int id)
{int m=(l+r)/2,ans=0;if(l==r)return 0;if(id<=m){ans=findans(l,m,e*2,id);if(ans<tree[e*2+1])ans=tree[e*2+1];return ans;}elsereturn findans(m+1,r,e*2+1,id);
}
int cmp(int aa,int bb)
{return aa<bb;
}
void cut()
{tn=1;sort(b+1,b+1+n,cmp);for(int i=2;i<=n;i++)if(b[i]!=b[tn])b[++tn]=b[i];
}
int two(int aa)
{int l,r,m;l=1;r=tn;while(l<=r){m=(l+r)/2;if(b[m]==aa)return m;if(b[m]>aa)r=m-1;else l=m+1;}return m;
}
int main()
{while(scanf("%d",&n)>0){if(n==0){printf("0\n");continue;}for(int i=1;i<=n;i++){scanf("%d",&a[i]);b[i]=a[i];}cut();builde(1,tn,1);int m=1;kuai[m].l=kuai[m].r=1;for(int i=2;i<=n;i++)if(a[kuai[m].r]>=a[i]){m++; kuai[m].l=kuai[m].r=i;}elsekuai[m].r++;for(int i=1;i<=m;i++){for(int j=kuai[i].r;j>=kuai[i].l;j--)k[j]=kuai[i].r-j+1;}int ans=0;for(int i=m;i>0;i--){int aaa,id;for(int j=kuai[i].r;j>=kuai[i].l;j--){id=two(a[j]);aaa=j-kuai[i].l+1;aaa+=findans(1,tn,1,id);if(ans<aaa)ans=aaa;}for(int j=kuai[i].r;j>=kuai[i].l;j--){id=two(a[j]);settree(1,tn,1,id,k[j]);}}printf("%d\n",ans);}
}/**************************************************************Problem: 1551User: aking2015Language: C++Result: AcceptedTime:192 msMemory:1320 kb
****************************************************************/

Problem F: Friends

Time Limit: 3 Sec  Memory Limit: 256 MB
Submit: 131  Solved: 25
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

3
2
2
34
2
5
3
8

Sample Output

1
2

HINT

二分匹配

#include<stdio.h>
#include<math.h>
#include <algorithm>
#include<string.h>
#include <time.h>
using namespace std;
#define ll long long
//-----大素数的判断-------
ll mult_mod(ll a,ll b,ll n)
{ll s=0;while(b){if(b&1) s=(s+a)%n;a=(a+a)%n;b>>=1;}return s;
}ll pow_mod(ll a,ll b,ll n)
{ll s=1;while(b){if(b&1) s=mult_mod(s,a,n);a=mult_mod(a,a,n);b>>=1;}return s;
}int Prime(ll n)
{ll u=n-1,pre,x;int i,j,k=0;if(n==2||n==3||n==5||n==7||n==11)  return 1;if(n==1||(!(n%2))||(!(n%3))||(!(n%5))||(!(n%7))||(!(n%11)))   return 0;for(;!(u&1);k++,u>>=1);srand((ll)time(0));for(i=0;i<5;i++){x=rand()%(n-2)+2;x=pow_mod(x,u,n);pre=x;for(j=0;j<k;j++){x=mult_mod(x,x,n);if(x==1&&pre!=1&&pre!=(n-1))return 0;pre=x;}if(x!=1)  return 0;}return 1;
}
int n,ok[105][105],mark[105],vist[105];
int DFS(int x)
{for(int i=1;i<=n;i++)if(ok[x][i]&&vist[i]==0){vist[i]=1;if(mark[i]==0||DFS(mark[i])){mark[i]=x; return 1;}}return 0;
}
int main()
{ll a[105];while(scanf("%d",&n)>0){for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}memset(ok,0,sizeof(ok));for(int i=1;i<n;i++)for(int j=i+1;j<=n;j++)if(Prime(a[i]+a[j]))ok[i][j]=ok[j][i]=1;int ans=0;memset(mark,0,sizeof(mark));for(int i=1;i<=n;i++){memset(vist,0,sizeof(vist));ans+=DFS(i);}printf("%d\n",ans/2);}
}/**************************************************************Problem: 1552User: aking2015Language: C++Result: AcceptedTime:1640 msMemory:1012 kb
****************************************************************/

Problem G: Good subsequence

Time Limit: 2 Sec  Memory Limit: 256 MB
Submit: 208  Solved: 48
[Submit][Status][Web Board]

Description

Input

Output

Sample Input

5 2
5 4 2 3 1
1 1
1

Sample Output

3
1

HINT

题意：最长连续子序列，要求子序列中最大减去最小的数小于k

思路：。。。。。。

转载请注明出处：寻找&星空の孩子

#include <stdio.h>
#include <stdio.h>
#include <algorithm>
#include <set>
using namespace std;set<int> st;
int a[10005];int main()
{int n,k,i,j,ans;while(~scanf("%d%d",&n,&k)){st.clear();for(i = 1;i<=n;i++)scanf("%d",&a[i]);ans = 0;for(i = 1,j = 1;i<=n;i++){st.insert(a[i]);while(*st.rbegin()-*st.begin()>k){st.erase(st.find(a[j]));j++;}ans = max(ans,i-j+1);}printf("%d\n",ans);}return 0;
}/**************************************************************Problem: 1553User: aking2015Language: C++Result: AcceptedTime:652 msMemory:1104 kb
****************************************************************/

The SG value of a set (multiset) is the minimum positive integer that could not be constituted of the number in this set.
You will be start with an empty set, now there are two opertions:
1. insert a number x into the set;
2. query the SG value of current set.

Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1e5) -- the total number of opertions.
The next N lines contain one opertion each.
1 x means insert a namber x into the set;
2 means query the SG value of current set.

For each query output the SG value of current set.

For sequence i1, i2, i3, … , iN, we set aj to be the number of members in the sequence which are prior to j and greater to j at the same time. The sequence a1, a2, a3, … , aN is referred to as the inversion sequence of the original sequence (i1, i2, i3, … , iN). For example, sequence 1, 2, 0, 1, 0 is the inversion sequence of sequence 3, 1, 5, 2, 4. Your task is to find a full permutation of 1~N that is an original sequence of a given inversion sequence. If there is no permutation meets the conditions please output “No solution”.

There are several test cases.
Each test case contains 1 positive integers N in the first line.(1 ≤ N ≤ 10000).
Followed in the next line is an inversion sequence a1, a2, a3, … , aN (0 ≤ aj < N)
The input will finish with the end of file.

For each case, please output the permutation of 1~N in one line. If there is no permutation meets the conditions, please output “No solution”.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
#define ls 2*i
#define rs 2*i+1
const int L = 10005;
struct node
{int l,r,n;
} a[L*4];
int n,ans[L],pos;void push_up(int i)
{a[i].n = a[ls].n+a[rs].n;
}void init(int l,int r,int i)
{a[i].l = l;a[i].r = r;if(l==r){a[i].n = 1;return ;}int mid = (l+r)/2;init(l,mid,ls);init(mid+1,r,rs);push_up(i);
}void query(int i,int x)
{if(a[i].l == a[i].r){pos = a[i].l;a[i].n = 0;return ;}if(x<=a[ls].n)query(ls,x);elsequery(rs,x-a[ls].n);push_up(i);
}int main()
{int i,j,k,x;while(~scanf("%d",&n)){memset(ans,0,sizeof(ans));init(1,n,1);int flag = 0;for(i = 1; i<=n; i++){scanf("%d",&x);if(flag)continue;if(a[1].n<x+1){flag = 1;continue;}else{query(1,x+1);ans[pos] = i;}}if(flag)printf("No solution\n");else{printf("%d",ans[1]);for(i = 2; i<=n; i++)printf(" %d",ans[i]);printf("\n");}}return 0;
}/**************************************************************Problem: 1555User: aking2015Language: C++Result: AcceptedTime:192 msMemory:1472 kb
****************************************************************/

Jerry is caught by Tom. He was penned up in one room with a door, which only can be opened by its code. The code is the answer of the sum of the sequence of number written on the door. The type of the sequence of number is

But Jerry’s mathematics is poor, help him to escape from the room.

There are some cases (about 500). For each case, there are two integer numbers n, m describe as above ( 1 <= n < 1 000 000, 1 <= m < 1000).

For each case, you program will output the answer of the sum of the sequence of number (mod 1e9+7).

#include<stdio.h>
#define LL long long
const LL mod=1e9+7;
inline LL two(LL &a,LL m)
{if(m==0)return 1;LL ans=two(a,m/2);ans=(ans*ans)%mod;if(m&1)ans=(ans*a)%mod;return ans;
}
int main()
{LL n,m,i;while(scanf("%lld%lld",&n,&m)>0){LL ans=0;for(i=1;i<=n;i++)ans=(ans+two(i,m))%mod;printf("%lld\n",ans);}
}/**************************************************************Problem: 1556User: aking2015Language: C++Result: AcceptedTime:3012 msMemory:964 kb
****************************************************************/