【84.62%】【codeforces 552A】Vanya and Table

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vanya has a table consisting of 100 rows, each row contains 100 cells. The rows are numbered by integers from 1 to 100 from bottom to top, the columns are numbered from 1 to 100 from left to right.

In this table, Vanya chose n rectangles with sides that go along borders of squares (some rectangles probably occur multiple times). After that for each cell of the table he counted the number of rectangles it belongs to and wrote this number into it. Now he wants to find the sum of values in all cells of the table and as the table is too large, he asks you to help him find the result.

Input
The first line contains integer n (1 ≤ n ≤ 100) — the number of rectangles.

Each of the following n lines contains four integers x1, y1, x2, y2 (1 ≤ x1 ≤ x2 ≤ 100, 1 ≤ y1 ≤ y2 ≤ 100), where x1 and y1 are the number of the column and row of the lower left cell and x2 and y2 are the number of the column and row of the upper right cell of a rectangle.

Output
In a single line print the sum of all values in the cells of the table.

Examples
input
2
1 1 2 3
2 2 3 3
output
10
input
2
1 1 3 3
1 1 3 3
output
18
Note
Note to the first sample test:

Values of the table in the first three rows and columns will be as follows:

121

121

110

So, the sum of values will be equal to 10.

Note to the second sample test:

Values of the table in the first three rows and columns will be as follows:

222

222

222

So, the sum of values will be equal to 18.

【题目链接】:http://codeforces.com/contest/552/problem/A

【题解】

枚举每个矩形占据了哪些格子;
用个数组递增那些位置的值就好了.

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 100+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int a[MAXN][MAXN];
int n;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
    {
        int a1,b1,a2,b2;
        scanf("%d%d%d%d",&a1,&b1,&a2,&b2);
        rep1(j,b1,b2)
            rep1(k,a1,a2)
                a[j][k]++;
    }
    LL ans = 0;
    rep1(i,1,100)
        rep1(j,1,100)
            ans+=a[i][j];
    cout << ans << endl;
    return 0;
}

转载于:https://www.cnblogs.com/AWCXV/p/7626832.html