【模拟】NEERC15 A Adjustment Office (Codeforces GYM 100851)
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题目链接:
http://codeforces.com/gym/100851
题目大意:
一个N*N的矩阵A,Ai,j=i+j,Q次操作,每次分两种,R r取出第r行还未被取的所有数,并输出和。C c取出第c列还未被取出的所有数并输出和。
题目思路:
【模拟】
首先Ai,j=i+j这个很关键。预处理每一行(=列)的值。
只要记录当前取得时候前面已经取过的所有行数的和、次数,所有列数的和、次数,就能推算出这次取数会少掉多少值。
并记录这一行或这一列被取过没有。
////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<stack>#include<queue>#include<set>#include<bitset>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-8)#define J 10#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#define N 1000004using namespace std;typedef long long LL;int cas,cass;int n,m,lll,ans;LL sum,tot,a,b;LL c[N];char s[2];bool u[N],v[N];int main(){#ifndef ONLINE_JUDGE//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;//for(scanf("%d",&cass);cass;cass--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s+1))while(~scanf("%d",&n)){sum=tot=a=b=0;mem(u,0);mem(v,0);scanf("%d",&m);c[1]=1LL*(2+1+n)*n/2;for(i=2;i<=n;i++)c[i]=c[i-1]+n;for(i=1;i<=m;i++){scanf("%s%d",s,&cas);if(s[0]=='R'){if(u[cas]){puts("0");continue;}printf("%I64d\n",c[cas]-tot-b*cas);sum+=cas;a++;u[cas]=1;}else{if(v[cas]){puts("0");continue;}printf("%I64d\n",c[cas]-sum-a*cas);tot+=cas;b++;v[cas]=1;}}}return 0;}/*////*/
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