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;

}

巴特西

Gym 100851A Adjustment Office (思维)

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