2016ACM/ICPC亚洲区大连站-重现赛

题目链接：http://acm.hdu.edu.cn/search.php?field=problem&key=2016ACM%2FICPC%D1%C7%D6%DE%C7%F8%B4%F3%C1%AC%D5%BE-%D6%D8%CF%D6%C8%FC%A3%A8%B8%D0%D0%BB%B4%F3%C1%AC%BA%A3%CA%C2%B4%F3%D1%A7%A3%A9&source=1&searchmode=source

A.染色乱搞。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef struct Edge {
 5     int to, next;
 6 }Edge;
 7 
 8 const int maxn = 1010;
 9 int n, m, x, y;
10 int know[maxn], vis[maxn], color[maxn];
11 int ecnt, head[maxn];
12 Edge e[maxn*maxn];
13 
14 void init() {
15     ecnt = 0;
16     memset(head, -1, sizeof(head));
17     memset(know, 0, sizeof(know));
18     memset(vis, 0, sizeof(vis));
19     memset(color, 0, sizeof(color));
20 }
21 
22 void adde(int u, int v) {
23     e[ecnt].to = v, e[ecnt].next = head[u];
24     head[u] = ecnt++;
25 }
26 
27 bool dfs(int u) {
28     for(int i = head[u]; ~i; i=e[i].next) {
29         int v = e[i].to;
30         if(color[v] == color[u]) return 0;
31         if(vis[v]) continue;
32         vis[v] = 1;
33         color[v] = 3 - color[u];
34         if(!dfs(v)) return 0;
35     }
36     return 1;
37 }
38 
39 int main() {
40     // freopen("in", "r", stdin);
41     int u, v;
42     while(~scanf("%d%d%d%d",&n,&m,&x,&y)) {
43         init();
44         for(int i = 0; i < m; i++) {
45             scanf("%d%d",&u,&v);
46             adde(u, v); adde(v, u);
47             know[u] = know[v] = 1;
48         }
49         for(int i = 0; i < x; i++) {
50             scanf("%d", &u);
51             color[u] = 1; know[u] = 1;
52         }
53         for(int i = 0; i < y; i++) {
54             scanf("%d", &u);
55             color[u] = 2; know[u] = 1;
56         }
57         bool flag = 0;
58         for(int i = 1; i <= n; i++) {
59             if(!know[i]) {
60                 flag = 1;
61                 break;
62             }
63         }
64         if(flag) {
65             puts("NO");
66             continue;
67         }
68         for(int i = 1; i <= n; i++) {
69             if(color[i]) {
70                 vis[i] = 1;
71                 if(!dfs(i)) {
72                     flag = 1;
73                     break;
74                 }
75             }
76         }
77         for(int i = 1; i <= n; i++) {
78             if(!color[i]) {
79                 vis[i] = 1;
80                 color[i] = 1;
81                 if(!dfs(i)) {
82                     flag = 1;
83                     break;
84                 }
85             }
86         }
87         if(!flag) puts("YES");
88         else puts("NO");
89     }
90     return 0;
91 }
A

C.威佐夫博弈+大数，根号5用mathmatica跑出来的，其实可以用二分打表。

 1 import java.math.BigDecimal;
 2 import java.math.BigInteger;
 3 import java.math.MathContext;
 4 import java.math.RoundingMode;
 5 import java.util.Scanner;
 6 
 7 public class Main {
 8 
 9     public static void main(String[] args) {
10         Scanner in = new Scanner(System.in);
11         BigDecimal n, m;
12         BigDecimal sqrt5 = new BigDecimal("2.2360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081761507737835");
13         while(in.hasNextBigDecimal()) {
14             n = in.nextBigDecimal(); m = in.nextBigDecimal();
15             if(m.equals(n.max(m))) {
16                 BigDecimal x = n;
17                 n = m;
18                 m = x;
19             }
20             BigDecimal k = n.subtract(m);
21             BigDecimal p = new BigDecimal(1);
22             n = p.add(sqrt5).multiply(k);
23             n = n.divide(new BigDecimal(2));
24             if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
25             else System.out.println("1");
26         }
27     }
28 }
C

二分打sqrt(5)

 1 import java.math.BigDecimal;
 2 import java.math.BigInteger;
 3 import java.math.MathContext;
 4 import java.math.RoundingMode;
 5 import java.util.Scanner;
 6 
 7 public class Main {
 8         public static BigDecimal Get(BigDecimal x) {
 9             BigDecimal lo = new BigDecimal("0.000000000000000000000000000001");
10             BigDecimal hi = x;
11             for(int i = 0; i < 1000; i++) {
12                 BigDecimal mid = lo.add(hi).divide(new BigDecimal("2.0"));
13                 BigDecimal mid2 = mid.multiply(mid);
14                 if(mid2.max(x).equals(mid2)) hi = mid;
15                 else lo = mid;
16             }
17             return hi;
18         }
19     public static void main(String[] args) {
20         Scanner in = new Scanner(System.in);
21         BigDecimal n, m;
22         BigDecimal sqrt5 = Get(new BigDecimal(5));
23         while(in.hasNextBigDecimal()) {
24             n = in.nextBigDecimal(); m = in.nextBigDecimal();
25             if(m.equals(n.max(m))) {
26                 BigDecimal x = n;
27                 n = m;
28                 m = x;
29             }
30             BigDecimal k = n.subtract(m);
31             BigDecimal p = new BigDecimal(1);
32             n = p.add(sqrt5).multiply(k);
33             n = n.divide(new BigDecimal(2));
34             if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
35             else System.out.println("1");
36         }
37     }
38 }
C

D.找到规律gcd(a,b)=gcd(x,y)以后，就是解方程了。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 LL a, b, x, y;
 6 
 7 LL gcd(LL x, LL y) {
 8     return y == 0 ? x : gcd(y, x % y);
 9 }
10 int main() {
11     // freopen("in", "r", stdin);
12     while(~scanf("%I64d%I64d",&a,&b)) {
13         LL g = gcd(a, b);
14         LL k = b * g;
15         if(a * a - 4 * k < 0 || (LL)sqrt(a*a-4*k) != sqrt(a*a-4*k)) puts("No Solution");
16         else {
17             LL delta = (LL)sqrt(a * a - 4 * k);
18             printf("%I64d %I64d
", (a-delta)/2, (a+delta)/2);
19         }
20     }
21     return 0;
22 }
D

F.考虑让整个数列a最长，那么就是公差为1的等差数列，这时候一定会有不够或者超过x的情况。这时候就把多余的部分从后往前+1补齐。问题就变成在这样一个数列里找到最长的不大于x的个数。由于有取模，所以除法用逆元来做。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 LL exgcd(LL a, LL b, LL &x, LL &y) {
 7     if(b == 0) {
 8         x = 1;
 9         y = 0;
10         return a;
11     }
12     else {
13         LL ret = exgcd(b, a%b, x, y);
14         LL tmp = x;
15         x = y;
16         y = tmp - a / b * y;
17         return ret;
18     }
19 }
20 
21 LL inv(LL a, LL m) {
22     LL x, y;
23     exgcd(a, m, x, y);
24     return (x % m + m) % m;
25 }
26 
27 const LL mod = (LL)1e9+7;
28 const int maxn = 60600;
29 LL f[maxn];
30 int x;
31 
32 void init() {
33     memset(f, 0, sizeof(f));
34     f[0] = 1; f[1] = 1;
35     for(int i = 2; i < maxn; i++) {
36         f[i] = (f[i-1] * i) % mod;
37     }
38 }
39 
40 int main() {
41     // freopen("in", "r", stdin);
42     init();
43     int T;
44     scanf("%d", &T);
45     while(T--) {
46         scanf("%d", &x);
47         if(x <= 1) {
48             printf("%d
", x);
49             continue;
50         }
51         int lo = 2, hi = maxn;
52         int pos = 2;
53         while(lo <= hi) {
54             int mid = (lo + hi) >> 1;
55             LL s = (mid + 2) * (mid - 1) / 2;
56             if(s <= x) {
57                 lo = mid + 1;
58                 pos = max(pos, mid);
59             }
60             else hi = mid - 1;
61         }
62         x -= (pos + 2) * (pos - 1) / 2;
63         if(x == pos) printf("%I64d
", (f[pos]*inv(2,mod))%mod*(pos+2)%mod);
64         else if(pos == 0) printf("%I64d
", f[pos]);
65         else printf("%I64d
", f[pos+1]*inv(pos-x+1,mod)%mod);
66     }
67     return 0;
68 }
F

H.奇数就无所谓，偶数的话先手有优势。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int k;
 5 
 6 int main() {
 7     // freopen("in", "r", stdin);
 8     while(~scanf("%d", &k)) {
 9         if(k & 1) puts("0");
10         else puts("1");
11     }
12     return 0;
13 }
H

I.余弦定理求出第三条边，高用角的一半的余弦乘一下斜边，面积加起来。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 11;
 5 const double pi = 3.141592653;
 6 int n, d;
 7 int th;
 8 double ret;
 9 
10 double f(int th) {
11     return d*cos(pi*th/2.0/180.0)*sqrt(2.0*d*d-2.0*d*d*cos(pi*th/180.0))/2.0;
12 }
13 
14 int main() {
15     // freopen("in", "r", stdin);
16     while(~scanf("%d%d",&n,&d)) {
17         ret = 0.0;
18         for(int i = 1; i <= n; i++) {
19             scanf("%d", &th);
20             ret += f(th);
21         }
22         printf("%.3lf
", ret);
23     }
24     return 0;
25 }
I

J.拆成四部分挨个看看是不是97。

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 const int maxn = 110;
 5 int n;
 6 int a;
 7 
 8 int f(int x) {
 9     int cnt = 0;
10     int q = 4;
11     while(q--) {
12         if((x % 256) == 97) cnt++;
13             x >>= 8;
14     }
15     return cnt;
16 }
17 
18 int main() {
19     // freopen("in", "r", stdin);
20     while(~scanf("%d", &n)) {
21         int ret = 0;
22         for(int i = 1;  i <= n; i++) {
23             scanf("%d", &a);
24             ret += f(a);
25         }
26         printf("%d
", ret);
27     }
28     return 0;
29 }
J
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原文地址：https://www.cnblogs.com/kirai/p/6035874.html