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。

 #include <bits/stdc++.h>
 using namespace std;
 
 typedef struct Edge {
     int to, next;
 }Edge;
 
 const int maxn = ;
 int n, m, x, y;
 int know[maxn], vis[maxn], color[maxn];
 int ecnt, head[maxn];
 Edge e[maxn*maxn];
 
 void init() {
     ecnt = ;
     memset(head, -, sizeof(head));
     memset(know, , sizeof(know));
     memset(vis, , sizeof(vis));
     memset(color, , sizeof(color));
 }
 
 void adde(int u, int v) {
     e[ecnt].to = v, e[ecnt].next = head[u];
     head[u] = ecnt++;
 }
 
 bool dfs(int u) {
     for(int i = head[u]; ~i; i=e[i].next) {
         int v = e[i].to;
         if(color[v] == color[u]) return ;
         if(vis[v]) continue;
         vis[v] = ;
         color[v] =  - color[u];
         if(!dfs(v)) return ;
     }
     return ;
 }
 
 int main() {
     // freopen("in", "r", stdin);
     int u, v;
     while(~scanf("%d%d%d%d",&n,&m,&x,&y)) {
         init();
         for(int i = ; i < m; i++) {
             scanf("%d%d",&u,&v);
             adde(u, v); adde(v, u);
             know[u] = know[v] = ;
         }
         for(int i = ; i < x; i++) {
             scanf("%d", &u);
             color[u] = ; know[u] = ;
         }
         for(int i = ; i < y; i++) {
             scanf("%d", &u);
             color[u] = ; know[u] = ;
         }
         bool flag = ;
         for(int i = ; i <= n; i++) {
             if(!know[i]) {
                 flag = ;
                 break;
             }
         }
         if(flag) {
             puts("NO");
             continue;
         }
         for(int i = ; i <= n; i++) {
             if(color[i]) {
                 vis[i] = ;
                 if(!dfs(i)) {
                     flag = ;
                     break;
                 }
             }
         }
         for(int i = ; i <= n; i++) {
             if(!color[i]) {
                 vis[i] = ;
                 color[i] = ;
                 if(!dfs(i)) {
                     flag = ;
                     break;
                 }
             }
         }
         if(!flag) puts("YES");
         else puts("NO");
     }
     return ;
 }

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。

 import java.math.BigDecimal;
 import java.math.BigInteger;
 import java.math.MathContext;
 import java.math.RoundingMode;
 import java.util.Scanner;
 
 public class Main {
 
     public static void main(String[] args) {
         Scanner in = new Scanner(System.in);
         BigDecimal n, m;
         BigDecimal sqrt5 = new BigDecimal("2.2360679774997896964091736687312762354406183596115257242708972454105209256378048994144144083787822749695081761507737835");
         while(in.hasNextBigDecimal()) {
             n = in.nextBigDecimal(); m = in.nextBigDecimal();
             if(m.equals(n.max(m))) {
                 BigDecimal x = n;
                 n = m;
                 m = x;
             }
             BigDecimal k = n.subtract(m);
             BigDecimal p = new BigDecimal(1);
             n = p.add(sqrt5).multiply(k);
             n = n.divide(new BigDecimal(2));
             if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
             else System.out.println("1");
         }
     }
 }

C

二分打sqrt(5)

 import java.math.BigDecimal;
 import java.math.BigInteger;
 import java.math.MathContext;
 import java.math.RoundingMode;
 import java.util.Scanner;
 
 public class Main {
         public static BigDecimal Get(BigDecimal x) {
             BigDecimal lo = new BigDecimal("0.000000000000000000000000000001");
             BigDecimal hi = x;
             for(int i = 0; i < 1000; i++) {
                 BigDecimal mid = lo.add(hi).divide(new BigDecimal("2.0"));
                 BigDecimal mid2 = mid.multiply(mid);
                 if(mid2.max(x).equals(mid2)) hi = mid;
                 else lo = mid;
             }
             return hi;
         }
     public static void main(String[] args) {
         Scanner in = new Scanner(System.in);
         BigDecimal n, m;
         BigDecimal sqrt5 = Get(new BigDecimal(5));
         while(in.hasNextBigDecimal()) {
             n = in.nextBigDecimal(); m = in.nextBigDecimal();
             if(m.equals(n.max(m))) {
                 BigDecimal x = n;
                 n = m;
                 m = x;
             }
             BigDecimal k = n.subtract(m);
             BigDecimal p = new BigDecimal(1);
             n = p.add(sqrt5).multiply(k);
             n = n.divide(new BigDecimal(2));
             if(n.toBigInteger().equals(m.toBigInteger())) System.out.println("0");
             else System.out.println("1");
         }
     }
 }

C

D.找到规律gcd(a,b)=gcd(x,y)以后，就是解方程了。

 #include <bits/stdc++.h>
 using namespace std;
 
 typedef long long LL;
 LL a, b, x, y;
 
 LL gcd(LL x, LL y) {
     return y ==  ? x : gcd(y, x % y);
 }
 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%I64d%I64d",&a,&b)) {
         LL g = gcd(a, b);
         LL k = b * g;
         if(a * a -  * k <  || (LL)sqrt(a*a-*k) != sqrt(a*a-*k)) puts("No Solution");
         else {
             LL delta = (LL)sqrt(a * a -  * k);
             printf("%I64d %I64d\n", (a-delta)/, (a+delta)/);
         }
     }
     return ;
 }

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。

 #include <bits/stdc++.h>
 using namespace std;
 
 typedef long long LL;
 
 LL exgcd(LL a, LL b, LL &x, LL &y) {
     if(b == ) {
         x = ;
         y = ;
         return a;
     }
     else {
         LL ret = exgcd(b, a%b, x, y);
         LL tmp = x;
         x = y;
         y = tmp - a / b * y;
         return ret;
     }
 }
 
 LL inv(LL a, LL m) {
     LL x, y;
     exgcd(a, m, x, y);
     return (x % m + m) % m;
 }
 
 const LL mod = (LL)1e9+;
 const int maxn = ;
 LL f[maxn];
 int x;
 
 void init() {
     memset(f, , sizeof(f));
     f[] = ; f[] = ;
     for(int i = ; i < maxn; i++) {
         f[i] = (f[i-] * i) % mod;
     }
 }
 
 int main() {
     // freopen("in", "r", stdin);
     init();
     int T;
     scanf("%d", &T);
     while(T--) {
         scanf("%d", &x);
         if(x <= ) {
             printf("%d\n", x);
             continue;
         }
         int lo = , hi = maxn;
         int pos = ;
         while(lo <= hi) {
             int mid = (lo + hi) >> ;
             LL s = (mid + ) * (mid - ) / ;
             if(s <= x) {
                 lo = mid + ;
                 pos = max(pos, mid);
             }
             else hi = mid - ;
         }
         x -= (pos + ) * (pos - ) / ;
         if(x == pos) printf("%I64d\n", (f[pos]*inv(,mod))%mod*(pos+)%mod);
         else if(pos == ) printf("%I64d\n", f[pos]);
         else printf("%I64d\n", f[pos+]*inv(pos-x+,mod)%mod);
     }
     return ;
 }

F

H.奇数就无所谓，偶数的话先手有优势。

 #include <bits/stdc++.h>
 using namespace std;
 
 int k;
 
 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d", &k)) {
         if(k & ) puts("");
         else puts("");
     }
     return ;
 }

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。

 #include <bits/stdc++.h>
 using namespace std;
 
 const int maxn = ;
 const double pi = 3.141592653;
 int n, d;
 int th;
 double ret;
 
 double f(int th) {
     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;
 }
 
 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d%d",&n,&d)) {
         ret = 0.0;
         for(int i = ; i <= n; i++) {
             scanf("%d", &th);
             ret += f(th);
         }
         printf("%.3lf\n", ret);
     }
     return ;
 }

I

J.拆成四部分挨个看看是不是97。

 #include <bits/stdc++.h>
 using namespace std;
 
 const int maxn = ;
 int n;
 int a;
 
 int f(int x) {
     int cnt = ;
     int q = ;
     while(q--) {
         if((x % ) == ) cnt++;
             x >>= ;
     }
     return cnt;
 }
 
 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d", &n)) {
         int ret = ;
         for(int i = ;  i <= n; i++) {
             scanf("%d", &a);
             ret += f(a);
         }
         printf("%d\n", ret);
     }
     return ;
 }

J