题目1 : Farthest Point

样例输入

1.000 1.000 5.000

样例输出

6 1

#include<bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int INF = 0x3f3f3f3f;
int dcmp(double x){if(fabs(x) < eps)return 0;return x < 0 ? -1 : 1;
}
int main(){double x, y, r;int left, right;double ansdis = -1;int ansx, ansy;cin >> x>> y >>r;left = ceil(x - r);right = floor(x + r);for(int i = right; i >= left; i--){int nowx = i;int high = floor(y + r),low = ceil(y),mid;int nowy = -INF;while(low <= high){mid = (low + high) >> 1;if(dcmp((mid-y)*(mid-y)+(i-x)*(i-x) - r* r) <= 0){nowy = mid;low = mid + 1;}else high = mid - 1;}if(nowy != -INF){double nowdis = (nowx-x)*(nowx-x)+(nowy-y)*(nowy-y);if(dcmp(nowdis - ansdis) > 0){ansx = i;   ansy = nowy;ansdis = nowdis;}}nowy = -INF;high = floor(y),low = ceil(y-r);while(low <= high){mid = (low + high) >> 1;if(dcmp((mid-y)*(mid-y)+(i-x)*(i-x) - r* r) <= 0){nowy = mid;high = mid - 1;}else low = mid + 1;}if(nowy != -INF){double nowdis = (nowx-x)*(nowx-x)+(nowy-y)*(nowy-y);if(dcmp(nowdis - ansdis) > 0){ansx = i;   ansy = nowy;ansdis = nowdis;}}}cout << ansx << " " << ansy << endl;return 0;
}

View Code

题目2 : Total Highway Distance

样例输入

3 5
1 2 2
2 3 3
QUERY
EDIT 1 2 4
QUERY
EDIT 2 3 2
QUERY

样例输出

10
14
12

全场最水题？给一棵树，以及边权，定义THD为点两两之间的距离和，一边修改一边查询THD的值思路：另其有根，设深度较深的Y，浅的为X， 则X必为Y的父亲。一条边链接X, Y，则其影响Y子树与外部的距离，即cnt[y]*(n-cnt[y])的贡献。处理出子树规模。记录每个点的深度及通向其父亲的边是哪一条。没了

#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
typedef long long ll;
struct Edge{int x, y, w, next;Edge(){};Edge(int u, int v, int z, int dis){x = u, y = v, next = z, w = dis;}
}edge[N<<1];int cntson[N], head[N], deep[N];
int faedge[N];int no;
ll ans;
void init(){memset(head, -1, sizeof(head));memset(cntson, 0, sizeof(cntson));no = 0;
}void add(int x, int y, int z){edge[no] = Edge(x, y , head[x], z);head[x] = no++;edge[no] = Edge(y, x, head[y], z);head[y] = no++;
}void dfs(int x, int fa){int i, y;cntson[x] = 1;for(i = head[x]; i != -1; i = edge[i].next){y = edge[i].y;if(y == fa)continue;deep[y] = deep[x] + 1;dfs(y, x);cntson[x] += cntson[y];faedge[y] = i;}
}
int main(){int n, m, x, y, z, i;char op[10];init();scanf("%d%d", &n, &m);for(i = 1; i < n; i++){scanf("%d%d%d", &x, &y, &z);add(x, y, z);}deep[1] = 0;dfs(1, 1);for(i = 0; i < no; i+=2){x = edge[i].x;  y = edge[i].y;if(deep[x] > deep[y])   swap(x, y);ans += (ll)(cntson[y])*(n - cntson[y])*edge[i].w;}while(m--){scanf("%s", op);if(op[0] == 'Q'){printf("%lld\n", ans);}else{scanf("%d%d%d",&x, &y, &z);if(deep[x] > deep[y])   swap(x, y);i = faedge[y];ans += (ll)(z - edge[i].w) * (cntson[y])*(n - cntson[y]);edge[i].w = z;}}return 0;
}

View Code

题目3 : Fibonacci

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

Given a sequence {an}, how many non-empty sub-sequence of it is a prefix of fibonacci sequence.

A sub-sequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

The fibonacci sequence is defined as below:

F1 = 1, F2 = 1

Fn = Fn-1 + Fn-2, n>=3

输入

One line with an integer n.

Second line with n integers, indicating the sequence {an}.

For 30% of the data, n<=10.

For 60% of the data, n<=1000.

For 100% of the data, n<=1000000, 0<=ai<=100000.

输出

One line with an integer, indicating the answer modulo 1,000,000,007.

样例提示

The 7 sub-sequences are:

{a2}

{a3}

{a2, a3}

{a2, a3, a4}

{a2, a3, a5}

{a2, a3, a4, a6}

{a2, a3, a5, a6}

样例输入

6
2 1 1 2 2 3

样例输出

7

处理出所有FIB值。对于Ai，若其为FIB，则求出她是第几个（fibj)，设cnt[j-1]为所有长度为j-1的符合题意的序列个数。则cnt[j] += cnt[j-1];特别处理一下x=1的时候。。

最后统计所有cnt就行了。注意取模= =

#include<bits/stdc++.h>
using namespace std;
const int N = 33;
const int MOD = 1000000007;typedef long long ll;int fib[N];
ll cnt[N];
bool vis[N];
int main(){int n, i,j, x;memset(fib, -1, sizeof(fib));memset(vis, 0, sizeof(vis));fib[1] = 1;fib[2] = 1;for(i = 3; fib[i-1] <= 100000; i++){fib[i] = fib[i-1] + fib[i-2];//printf("%d %d\n", i, fib[i]);
    }//fib[25]//1!!!!!!!!!!!!!!
memset(cnt, 0, sizeof(cnt));ll ans = 0;scanf("%d", &n);for(i = 1; i <= n; i++){scanf("%d", &x);if(x == 1){cnt[2] =(cnt[1] + cnt[2]) % MOD;cnt[1] ++;if(cnt[2] >=1)  vis[2] = true;continue;}ll tmp = 1;for( j = 3; j <= 25 && fib[j] <= x; j++){if(x == fib[j]) break;}if(x == fib[j]) {if(vis[j-1]){cnt[j] = (cnt[j] + cnt[j-1]) % MOD;vis[j] = true;}}// cout << i << " " << ans<<endl;
    }for(i = 1; i <= 25; i++){ans = (ans + cnt[i]) % MOD;}cout << ans<<endl;return 0;
}

View Code

题目4 : Image Encryption

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

描述

A fancy square image encryption algorithm works as follow:

0. consider the image as an N x N matrix

1. choose an integer k∈ {0, 1, 2, 3}

2. rotate the square image k * 90 degree clockwise

3. if N is odd stop the encryption process

4. if N is even split the image into four equal sub-squares whose length is N / 2 and encrypt them recursively starting from step 0

Apparently different choices of the k serie result in different encrypted images. Given two images A and B, your task is to find out whether it is POSSIBLE that B is encrypted from A. B is possibly encrypted from A if there is a choice of k serie that encrypt A into B.

输入

Input may contains multiple testcases.

The first line of the input contains an integer T(1 <= T <= 10) which is the number of testcases.

The first line of each testcase is an integer N, the length of the side of the images A and B.

The following N lines each contain N integers, indicating the image A.

The next following N lines each contain N integers, indicating the image B.

For 20% of the data, 1 <= n <= 15

For 100% of the data, 1 <= n <= 100, 0 <= Aij, Bij <= 100000000

输出

For each testcase output Yes or No according to whether it is possible that B is encrypted from A.

样例输入

3
2
1 2
3 4
3 1
4 2
2
1 2
4 3
3 1
4 2
4
4 1 2 3
1 2 3 4
2 3 4 1
3 4 1 2
3 4 4 1
2 3 1 2
1 4 4 3
2 1 3 2

样例输出

Yes
No
Yes

赛时犯了一个极其脑残的错误导致样例都过不去。。。泪目。。。。。。。。。。T T暴力了一个写法，不知道能不能过。等题库放题了再改吧【update：能过】

#include <bits/stdc++.h>
using namespace std;const int N = 104;
int gox[4] = {1, 1, -1, -1};
int goy[4] = {1, -1, -1, 1};
struct Matrix{int f[N][N];int mat_size;void scan(){int i, j;for(i = 1; i <= mat_size; i++)for(j = 1; j <= mat_size; j++)scanf("%d", &f[i][j]);}void print(){int i, j;for(i = 1; i <= mat_size; i++){for(j = 1; j <= mat_size; j++)printf("%d ", &f[i][j]);printf("\n");}}void modify(int t, int ra, int rb, int ca, int cb){Matrix C;int i, j, ii, jj;if(t & 1){if(t == 1){for(j = cb, ii = ra; j >= ca; j--, ii++)for(i = ra, jj = ca; i <= rb; i++, jj++)C.f[ii][jj] = f[i][j];}else{for(j = ca, ii = ra; j <= cb; j++, ii++)for(i = rb, jj = ca; i >= ra; i--, jj++)C.f[ii][jj] = f[i][j];}}else{for(i = rb, ii = ra; i >= ra; i--, ii++)for(j = cb, jj = ca; j >= ca; j--, jj++)C.f[ii][jj] = f[i][j];}for(i = ra; i <= rb; i++)for(j = ca; j <= cb; j++)f[i][j] = C.f[i][j];}bool check(const Matrix &I, int ra, int rb, int ca, int cb)const{int i, j;for(i = ra; i <= rb; i++)for(j = ca; j <= cb; j++)if(f[i][j] != I.f[i][j])    return false;return true;}
}A, B;bool judge(Matrix &A, Matrix &B, int ra, int rb, int ca, int cb){int i;Matrix Tmp = A;for(i = 0; i < 4; i++){if(i != 0)A.modify(i, ra, rb, ca, cb);if(((ra - rb + 1) &1 )){if(B.check(A, ra, rb, ca, cb)) return true;}else{if(B.check(A, ra, rb, ca, cb))return true;if(judge(A, B, ra, (ra + rb) / 2, ca, (ca+cb)/2) &&judge(A, B, (ra + rb) / 2+1, rb, (ca + cb) / 2 + 1, cb)&&judge(A, B, ra, (ra + rb) / 2, (ca + cb) / 2 + 1, cb) &&judge(A, B, (ra + rb) / 2+1, rb, ca, (ca+cb)/2))return true;}A = Tmp;}return false;
}
int main(){int n, TC, i, j;scanf("%d", &TC);while(TC--){scanf("%d", &n);A.mat_size = B.mat_size = n;A.scan();B.scan();printf("%s\n", judge(A, B, 1, n, 1, n)?"Yes" : "No");//  A = A.modify(1, 1,n, 1, n);// A.print();
    }return 0;
}

View Code