Codeforces 835 C Star sky
来源:互联网 发布:哈曼卡顿和bose 知乎 编辑:程序博客网 时间:2024/05/29 12:27
The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c).
Over time the stars twinkle. At moment 0 the i-th star has brightness si. Let at moment t some star has brightness x. Then at moment (t + 1) this star will have brightness x + 1, if x + 1 ≤ c, and 0, otherwise.
You want to look at the sky q times. In the i-th time you will look at the moment ti and you will see a rectangle with sides parallel to the coordinate axes, the lower left corner has coordinates (x1i, y1i) and the upper right — (x2i, y2i). For each view, you want to know the total brightness of the stars lying in the viewed rectangle.
A star lies in a rectangle if it lies on its border or lies strictly inside it.
The first line contains three integers n, q, c (1 ≤ n, q ≤ 105, 1 ≤ c ≤ 10) — the number of the stars, the number of the views and the maximum brightness of the stars.
The next n lines contain the stars description. The i-th from these lines contains three integers xi, yi, si (1 ≤ xi, yi ≤ 100, 0 ≤ si ≤ c ≤ 10) — the coordinates of i-th star and its initial brightness.
The next q lines contain the views description. The i-th from these lines contains five integers ti, x1i, y1i, x2i, y2i (0 ≤ ti ≤ 109, 1 ≤ x1i < x2i ≤ 100, 1 ≤ y1i < y2i ≤ 100) — the moment of the i-th view and the coordinates of the viewed rectangle.
For each view print the total brightness of the viewed stars.
2 3 31 1 13 2 02 1 1 2 20 2 1 4 55 1 1 5 5
303
3 4 51 1 22 3 03 3 10 1 1 100 1001 2 2 4 42 2 1 4 71 50 50 51 51
3350
Let's consider the first example.
At the first view, you can see only the first star. At moment 2 its brightness is 3, so the answer is 3.
At the second view, you can see only the second star. At moment 0 its brightness is 0, so the answer is 0.
At the third view, you can see both stars. At moment 5 brightness of the first is 2, and brightness of the second is 1, so the answer is 3.
题目大意:
在一个笛卡尔坐标系中,有n颗星星,然后每颗星星都有自己的初始亮度,然后整个星系中都有最亮的亮度,星星的亮度会随着时间而增加,但是到达最亮值的下一个单位的时间就会变成0,然后继续。。。。题目让我们求在一个矩阵里的所有星星T秒后的亮度总和
解题思路:
先把图存下来,然后能解出其他秒数时候的图,之后求矩阵和,,注意,同一个点可能有两颗星星
AC代码:
#include<stdio.h>int a[201][201][20];int main(){ int n, m, k; scanf("%d%d%d", &n, &m, &k); for(int i = 0; i < n; i++){ int x, y, z; scanf("%d%d%d", &x, &y, &z); for(int j = 0; j <= k; j++){ a[x][y][j] += (z + j) % (k + 1); } } for(int i = 1; i <= 100; i ++){ for(int j = 1; j <= 100; j++){ for(int l = 0; l <= k; l++){ a[i][j][l] += a[i - 1][j][l] + a[i][j - 1][l] - a[i - 1][j - 1][l]; } } } int t, x1, y1, x2, y2; while(m--){ scanf("%d%d%d%d%d", &t, &x1, &y1, &x2, &y2); t %= (k + 1); printf("%d\n", a[x2][y2][t] - a[x2][y1 - 1][t] - a[x1 - 1][y2][t] + a[x1 - 1][y1 - 1][t]); }}
- Codeforces 835C-Star sky
- codeforces 835 C Star sky
- Codeforces 835 C Star sky
- CodeForces 835C Star sky
- codeforces 835c Star sky
- C. Star sky Codeforces
- Codeforces #835C: Star Sky 题解
- 【Codeforces 835 C. Star sky】+ dp
- CodeForces 835 C.Star sky(水~)
- codeforces 835C Star sky(二维树状数组)
- codeforces 835-C. Star sky(dp+前缀和)
- codeforces 835C Star sky (二维数组前缀和)
- Codeforces 835 C Star sky(前缀和)
- CF -- 835C Star sky 【dp + 预处理】
- CodeFroces 835C. Star sky(构造题)
- C. Star sky(Codeforces Round #427 (Div. 2) C)
- Codeforces 835C Star sky【思维+暴力预处理二维前缀和】
- Codeforces Round #427 (Div. 2)C. Star sky
- JavaScript 类型转换
- 【面经笔记】字节对齐
- 软工文档小结
- Spring源码-目录
- Sqlserver获取所有数据库名,表信息,字段信息,主键信息,以及表结构等。
- Codeforces 835 C Star sky
- 使用Maven的体验
- 数据仓库--数据仓库概述
- PHP 模拟post请求发送数据
- 2017/7/29 第六天
- 单点登录原理与简单实现
- http协议
- Android Studio升级到2.3.3之后的打包问题
- 集合(1)循环取集合中的值(+取特定字段的值)