Processing math: 100%

CCPC江西省赛

,
这是一个值得庆幸的比赛
## [1004 Wave](http://acm.hdu.edu.cn/showproblem.php?pid=6570)

思路

存下每种数字的位置，

AC代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
#include<iostream>
#include<cmath>
#include<stdio.h>
#include<bits/stdc++.h>
#define LL long long
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(3)
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 45.0 / atan(1.0);
vector<int> vec[105];
int main()
{
    fio;
    int n, c, x;
    cin >> n >> c;
    for(int i = 1; i <= n; ++i)
    {
        cin >> x;
        vec[x].push_back(i);//下标是数字，内容是位置
    }
    int ans = 0;
    for(int i = 1; i <= c; ++i)
    {
        for(int j = 1; j <= c; ++j)
        {
            if(i == j)
                continue;
            int len1 = vec[i].size();
            int len2 = vec[j].size();
            int st1 = 0, st2 = 0, temp = 0, sum = 0;
            while(1)
            {
                while(st1 < len1 && vec[i][st1] < temp)
                {
                    st1 += 1;
                }
                if(st1 == len1)
                    break;
                temp = vec[i][st1];
                sum += 1;
                while(st2 < len2 && vec[j][st2] < temp)
                {
                    st2 += 1;
                }
                if(st2 == len2)
                    break;
                sum += 1;
                temp = vec[j][st2];
            }
            ans = max(ans, sum);
        }
    }
    cout << ans << endl;
    return 0;
}

1008 Rng放一放!

思路

  1. 这是什么神仙思路，假设区间趋于0，则可以看成是两个点，计算不相交的情况，就是第一个区间n种,第二个区间(n - 1)种，左右算两遍就是
  2. 具体

AC代码

1

1010 Worker

思路

最小公倍数

AC代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
/*************************************************************************
 > FileName:
 > Author:      Lance
 > Mail:        lancelot_hcs@qq.com
 > Date:        9102.1.8
 > Description:
 ************************************************************************/
//#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")//add_stack
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <vector>
#include <cstdio>
#include <bitset>
#include <string>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const int eps = 1e-10;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;//int2147483647//ll9e18//unsigned ll 1e19
const int maxn = 100005;

ll read()
{
    ll x = 0,f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
        f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}

ll LCM(ll a, ll b)
{
    return a * b / __gcd(a, b);
}
ll t, n, m, a[maxn], b[maxn];
int main()
{
//    ios::sync_with_stdio(false);
    while (~scanf("%lld%lld", &n, &m))
    {
        ll temp = 1;
        ll sum = 0, sum1 = 0;
        for (int i = 1; i <= n; i++)
        {
            scanf("%lld", &a[i]);
            temp = LCM(a[i], temp);
            sum1 += a[i];
        }
        for (int i = 1; i <= n; i++)
        {
            b[i] = temp / a[i];
            sum += b[i];
        }
        if (!(m % sum))
        {
            printf("Yes\n");
            for (int i = 1; i <= n; i++)
            {
                printf(i == n ? "%lld\n" : "%lld ", b[i] * (m / sum));
            }
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}


1007 Traffic

思路

遍历使得任意a[i]!=b[j]+x,求最小的x;

AC代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
/*************************************************************************
 > FileName:
 > Author:      Lance
 > Mail:        lancelot_hcs@qq.com
 > Date:        9102.1.8
 > Description:
 ************************************************************************/
//#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:102400000,102400000")//add_stack
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <vector>
#include <cstdio>
#include <bitset>
#include <string>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <map>
#include <set>
using namespace std;
typedef long long ll;
const double pi = acos(-1.0);
const int eps = 1e-10;
const int mod = 1e9 + 7;
#define debug(a) cout << "*" << a << "*" << endl
const int INF = 0x3f3f3f3f;

const int maxn = 1e6 + 5;

ll read()
{
    ll x = 0,f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
int t, n, m, a[maxn], b[maxn], v[maxn];
int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        memset(v, 0, sizeof(v));
        for (int i = 0; i < n; i++)
            a[i] = read();
        for (int i = 0; i < m; i++)
            b[i] = read();
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (a[i] >= b[j])
                {
                    v[a[i] - b[j]] = 1;
                }
            }
        }
        for (int i = 0;; i++)
            if (!v[i])
            {
                printf("%d\n", i);
                break;
            }

    }
    return 0;
}