UVALive 6852 Blanket（暴力）

题意:

$给出N\le 10^5个地毯, M\le 10^6个人, 人的编号是0\to M-1$
$地毯由二元组描述,(a,b), a表示[0,a]是厚的, b为地毯的长度周期, 地毯无限长$
$现在N个地毯摞在, 人排一行, 求被0\to N个地毯厚的部分覆盖的人数$

分析:

$题目数据范围有误, 实测应该是0\le a < b \le 16$
$记录cnt[T][mod]:= 周期为T余数为mod的厚部分毛毯个数$
$然后枚举一遍[0,M)暴力即可$

代码:

//
//  Created by TaoSama on 2015-12-12
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
#include <cassert>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, m, ans[N], cnt[20][20];

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d%d", &n, &m);
        memset(cnt, 0, sizeof cnt);
        memset(ans, 0, sizeof ans);
        for(int i = 1; i <= n; ++i) {
            int a, b; scanf("%d%d", &a, &b);
            for(int j = 0; j < a; ++j) ++cnt[b][j];
        }
        for(int i = 0; i < m; ++i) {
            int cur = 0;
            for(int j = 1; j <= 16; ++j)
                cur += cnt[j][i % j];
            ++ans[cur];
        }
        for(int i = 0; i <= n; ++i) printf("%d\n", ans[i]);
    }
    return 0;
}