Gym 100851A Adjustment Office (思维)

题意：给定一个 n*n 的矩阵，然后有 m 个询问，问你每一行或者每一列总是多少，并把这一行清空。

析：这个题不仔细想想，还真不好想，我们可以根据这个题意，知道每一行或者每一列都可以求和公式来求，然后再送去变成0的数，由于每一行或者每一列，

都是等差数列，所以我们只要记录每一个的第一个元素就好，再记录有多少个，然后就可以推算出来。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
char s[5];
bool row[maxn], col[maxn];

int main(){
    freopen("adjustment.in", "r", stdin);
    freopen("adjustment.out", "w", stdout);
    while(scanf("%d %d", &n, &m) == 2){
        memset(row, false, sizeof row);
        memset(col, false, sizeof col);
        int x = 0;
        LL r = 0, c = 0;
        int cntr = 0, cntc = 0;
        LL ans = 0;
        for(int i = 0; i < m; ++i){
            scanf("%s %d", s, &x);
            if(s[0] == ‘R‘){
                if(row[x])  ans = 0;
                else{
                    r += x;
                    ++cntr;
                    ans = (LL)n*(LL)(x+x+n+1)/2LL - (LL)cntc*x - c;
                }
                row[x] = true;
            }
            else {
                if(col[x]) ans = 0;
                else{
                    c += x;
                    ++cntc;
                    ans = (LL)n*(LL)(x+x+n+1)/2LL - (LL)cntr*x - r;
                }
                col[x] = true;
            }
            printf("%I64d\n", ans);
        }
    }
    return 0;
}