Codeforces 295A. Greg and Array Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li +...
Greg and Array CodeForces - 296C - 差分 给定一个数列 a=a1,a2,...,an 以及m 次操作,q 次查询。 其中第 i 次操作如同:li,ri,di,意指区间 [li,ri] 中每个元素加上 di。 其中第 i 次查询如同:xi,yi,意指需要执行第 xi,xi+1,...yi 次操作。 现在问你,执行完所有查询后的序列 a 是怎么样...
Greg has an array a = a1, a2, ..., an and m operations. Each operation looks as: li, ri, di, (1 ≤ li ≤ ri ≤ n). To apply operation i to the array means to increase all array elements with numbers li, li + 1, ..., ri by value ...
Hi everyone, Please help me finding out the mistake I did inGreg and Array. I am usingBinary Indexed Treeto solve this problem and Implemented the idea ofthis site(main idea is given below, however, please visit the site for complete documentation of the algorithm).. My complete code isAva...
CodeForces Round #179 (295A) - Greg and Array 一个线段树做两次用,次...先扫描1~k...用线段树统计出每个操作执行的次数..
C. Greg and Array time limit per test 1.5 seconds memory limit per test 256 megabytes Greg has an arraya = a1, a2, ..., anandmoperations. Each operation looks as:li,ri,di,(1 ≤ li ≤ ri ≤ n). To apply operationito the array means to increase all array...
1343B - Balanced Array.cpp Added 1343B - Balanced Array.cpp Mar 6, 2023 1345A - Puzzle Pieces.cpp Added 1345A - Puzzle Pieces.cpp Sep 24, 2023 1348A - Phoenix and Balance.cpp Added 1348A - Phoenix and Balance.cpp May 29, 2023 ...
#include<iostream>#include<algorithm>usingnamespacestd;typedeflonglongll;constintN=1e5+10;ll a[N],b[N];ll kk[N];structnode{intleft,right;ll max;ll lazy;}e[N*4];structop{intleft,right;ll val;}op[N];voidpush_up(intp){e[p].max=max(e[p*2].max,e[p*2+1].max);}voidpush_...
#280 Greg's Workout updated 10:56 pm 12.6.2024 Jun 13, 2024 #281 Find The Array updated 10:56 pm 12.6.2024 Jun 13, 2024 #282 Sofia and the Lost Operations updated 10:56 pm 12.6.2024 Jun 13, 2024 #283 Vlad and Candies updated 10:56 pm 12.6.2024 Jun 13, 2024 ...
B.Yaroslav and Two Strings(转载 原文地址) 对于两个字符串ch1和ch2,开四个数组a[i],b[i],c[i],d[i]分别表示 所有的情况数 、ch1[i]<=ch2[i]的情况数、ch1[i]>=ch2[i]的情况数、ch1[i]==ch2[i]的情况数,那么根据容斥原理,有ans = ∏a[i] - ∏b[i] - ∏c[i] + ...