}voidmodify(intx,intv) {for(inti = x; i < N; i += i & -i) a[i] +=v; }intsum(intx) {intans =0;for(inti = x; i; i -= i & -i) ans +=a[i];returnans; }intquery(intL,intR) {if(L > R)return0;returnsum(R) - sum(L -1); } } bit, bit2;intF[N], Finv...
CodeForces 1521E Nastia and a Beautiful Matrix 题解CF1521E链接这道题有多解,下面给出一种无需二分的做法。Hint 1 先假设已经求出了 $size$。 观察题目所给条件,是否可以把网格分成两部分,使得每部分里互不影响,只需要考虑两部分之间的冲突?Hint 2 按行的奇偶性来分,那么网格应当是这样的: 11111 2 ...
CodeForces 1521E Nastia and a Beautiful Matrix 题解 CF1521E链接这道题有多解,下面给出一种无需二分的做法。Hint 1 先假设已经求出了 sizesize。 观察题目所给条件,是否可以把网格分成两部分,使得每部分里互不影响,只需要考虑两部分之间的冲突?
Beautiful Matrix You've got a 5 × 5 matrix, consisting of 24 zeroes and a single number one. Let's index the matrix rows by numbers from 1 to 5 from top to b...A - Beautiful numbers CodeForces - 55D Volodya is an odd boy and his taste is strange as well. It seems ...
Codeforces 1085G Beautiful Matrix 树状数组,BeautifulMatrix首先枚举哪一位分出大小,然后计算比它小的方案数有多少种。方案数=当前行的方案数*下面所有行的方案数。如果下面还有x行,那么下面所有行的方案数为g^x,g为全错排的方案数,当前行的要预处理出f[i][j]f[i][j]
1365A-MatrixGame.cpp 1365B-TroubleSort.cpp 1365C-RotationMatching.cpp 1366A-ShovelsAndSwords.cpp 1366B-Shuffle.cpp 1366C-PalindromicPaths.cpp 1367A-ShortSubstrings.cpp 1367B-EvenArray.cpp 1367C-SocialDistance.cpp 1368A-C+eq.cpp 1368B-CodeforcesSubsequences.cpp 1368C-EvenPicture.cpp 1369A-Fashi...
we are inserting all vectors into a matrix, and get it down to row echelon form. Question 1: According to wikipedia, reduced row echelon form is unique. Does that imply that the bits we find will always be the same? I don't know :) I think so, but I'm not good at this. ...
Codeforces 1085G(1086E) Beautiful Matrix dpdp+树状数组 题意定义一个n∗nn∗n的矩阵是beautifulbeautiful的,需要满足以下三个条件:1.每一行是一个排列。2.上下相邻的两个元素的值不同。再定义两个矩阵的字典序大的矩阵大(从左往右从上到下一个一个比较)。给出一个beautifulbeautiful的n∗nn∗n的矩阵...
}usingio :: gi;usingio :: putc;usingio :: print;intmain(void){freopen("aztec.in","r", stdin),freopen("aztec.out","w", stdout);inttestCase =1;gi(testCase);while(testCase--) {intN, M, mx =0;gi(N),gi(M); std::vector <int>A(N +1),E(N +1,0);for(inti =1; i ...