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[题解] 2道几何学题目...BZOJ1041,1043 | sxysxy 's Blog

http://www.sxysxy.org/blogs/97

题解] 2道几何学题目.BZOJ1041,1043. 顺便一说有个神奇的函数叫atan2 (求向量与x轴的夹角 返回值 [-π，π]，要特判一下返回值的各种情况。 PI = acos(- 1. X, y; point(){} point( double. Y):x( x), y( y){} inline. Atan2(y, x); } inline. Dist(point &o) { return. Sqrt( x-o.x)*(x-o.x) (y-o.y)*(y-o.y) ; } point operator. Point &o) { return. Point(x-o.x, y-o.y); } }; struct. Circle { point o; double. R; circle(){} }cs[ 1001. L, r; range(){} range( double. R):l( l), r( r){} bool. L ol (l = o.l & r o.r); } };vector range rs; double. Dcmp(x-y); } void. Ar*d) ; double.

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时间线 | sxysxy 's Blog

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数学, 数论, 莫比乌斯反演. 题解] USACO cow hopscotch(牛跳房子). DP, 数学, CDQ分治, 数据结构. 模板] 后缀自动机- 树- 数组. Built with wblog(based on rails5).

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Built with wblog(based on rails5).

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[讲解] 一类"最大子矩阵和"问题的解法 | sxysxy 's Blog

http://www.sxysxy.org/blogs/94

暴力枚举子矩阵并求和n 6 前缀和优化n 5 naïve! R = - 0x6666666. J = n; j ) { scanf(. M[i][j]); m[i][j] = m[i][j- 1. J = i; j = n; j ) { memset(dp, 0. Dp) ; for. S = m[k][j]-m[k][i- 1. Dp[k] = max( 0. S; r = max(r, dp[k]); } } printf(. 可以对每个面做一个2维前缀和(也可以直接做3维前缀和，但是递推式稍麻烦.)，然后枚举 最大子空间 的顶面，向下做1维情形的最大子数组即可。 Z) { x = i& 1. X y z)% 2. LL; LL inf = 1. LL S[MAXN][MAXN][MAXN]; LL sum( int. Dx = x2-x1 1. Dy = y2-y1 1. Dz = z2-z1 1. I ) { int. X, y, z;expand(i, x, y, z); s -= S[x2-x*dx][y2-y*dy][z2-z*dz]*sign(x,y,z); } return.

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[记录] 《算法竞赛入门经典训练指南》1.5习题 | sxysxy 's Blog

http://www.sxysxy.org/blogs/92

记录] 算法竞赛入门经典训练指南 1.5习题. U = n) break. T ; } printf(. Kase , t); } return. A[MAXN], d[MAXN]; inline. X:-x; } bool. Sabs(a) sabs(b); } int. Main() { / freopen(test data.txt, r, stdin);. T- ) { int. I = n; i )scanf(. A i); std: sort(a 1. N, cmp); int. C); } return. 找规律: ans = n*m-1. M, n; while. M, &n) printf(. P tlength(); p ) if. S[q] = t[p])q ; puts(q = s.length()? 经典的线段覆盖问题，然而被坑在了 计算几何 强行计算几何系列 上. A[MAXN], vis[MAXV], num[MAXN]; vector int. N) & n){ if. Vis) ; for. I = n; i )vec[i].clear(); int.

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[题解] BZOJ2190&SDOI2008仪仗队 | sxysxy 's Blog

http://www.sxysxy.org/blogs/95

那么求一下 ans = phi(i), i = 2,3,.,n-1. 对称*2，再加上第一列，最后一行，主对角线上三个特殊点，即ans * 2 3就是最后的答案。 Pre() { phi[ 1. I = MAXN; i ) { if. Vis[i]) { phi[i] = i- 1. Prime[tot ] = i; } for. J tot; j ) { int. K = prime[j]*i; if. I%prime[j])phi[k] = phi[i]*phi[prime[j] ; / 积性. Phi[k] = phi[i]*prime[j]; / notice:if(i%p = 0)phi[i*p] = phi[i]*p. Main() { pre(); int. I n; i )ans = phi[i]; printf(. Ans = miu[n/i]*i; return. 发表于 2017.01.09. 自由转载 - 非商用 - 非衍生 - 保持署名. 题解] USACO cow hopscotch(牛跳房子). 模板] 后缀自动机- 树- 数组. 题解 讲解 模板] FFT 高精度乘法.

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[讲解] 一种多目标值的动态规划问题的解决方法 | sxysxy 's Blog

http://www.sxysxy.org/blogs/91

问题就可以转化为最小化x = Ma c。 X最小化时，a = x/M，c = x%M，显然此时a, c都最小化，问题得解。 Std; vector int. D[u][p]; vis[u][p] = 1. Ans = d[u][p]; ans = 2000. I G[u].size(); i ) if. F)ans = dp(G[u][i], u, 1. If root or 父节点已经放灯，u可以不放. I G[u].size(); i ) if. F)s = dp(G[u][i], u, 0. S ; ans = min(ans, s); } return. Ans; } int. Main() { / freopen(test data.txt, r, stdin);. T- ) { scanf(. N, &m); for. I n; i )G[i].clear(); for. X, y;scanf(. X, &y); G[x].push back(y); G[y].push back(x); } memset(vis, 0. Vis) ; int. R = dp(i, - 1.

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[题解] POI2014快递员 一道令人愉悦的好题！ | sxysxy 's Blog

http://www.sxysxy.org/blogs/93

Fs, *ft; inline. Fs= ft& (ft=(fs=buf) fread(buf, 1. Fs ; } inline. Fast read() { int. C = readc() { if. R = c 0x30. Isdigit(c = readc() ) r = (r 3. R:r; } } using. IO: fast read; #define. Arr[MAXN]; vector int. N, m; n = fast read(); m = fast read(); for. I = n; i )pos[arr[i] = fast read()].push back(i); while. M- ) { int. L = fast read(); int. R = fast read(); bool. Flag & t- ) { x = arr[rand()%(r-l 1. P = upper bound(pos[x].begin(), pos[x].end(), l- 1. Pos[x].begin(); int. Pos[x].begin(); if.

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[题解] COGS2569 博丽灵梦 梦想妙珠 | sxysxy 's Blog

http://www.sxysxy.org/blogs/96

题解] COGS2569 博丽灵梦 梦想妙珠. 一个数在数列中出现的位置一定是单调递增的(这很显然)，如果需要一个数[l, r]区间内出现的次数，只需要查询[l, r]里面有几个位置记录就好了。 Fs, *ft; inline. Fs= ft& (ft=(fs=buf) fread(buf, 1. Fs ; } inline. Fast read() { int. C = readc() { if. R = c 0x30. Isdigit(c = readc() ) r = (r 3. R:r; } } using. IO: fast read; #define. Main() { / freopen(test data.txt, r, stdin);. N = fast read(); for. I = n; i ) pos[fast read()].push back(i); int. Q = fast read(); while. Q- ) { int. L, r, v; l = fast read(); r = fast read(); v = fast read(); int. Node *x...

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