第13届 广工校赛 题解汇总

题目链接：13届广工校赛

貌似牛客网的算法群里已经有题解的pdf了，那我不复述那么多了。 贴贴AC代码就走~~

A-跳台阶

#include 
using namespace std;
 
int n;
long long a[40];
int main(){
    a[1] = 1;
    for (int i=2; i<=30; i++) {
        a[i] = 1;
        for (int j=1; j

B-跳一跳，很简单的

额，这题代码不是我自己写出来的，编程技巧上我还不够火候。仿照某位大神写出来的，多谢大神的代码share

#include 
#include 
#include 
#include 
using namespace std;

typedef long long LL;
const int maxn = 1e5+1;
const int maxc = 26;
const int mod = 1e9+7;
const int seed = 29;
const int maxd = 17;

int n,Q;
int a[maxn],b[maxn],pw[maxn];
int dep[maxn],fa[maxd][maxn];
int h[maxc+1][maxn];
vector e[maxn];


void init(){
   pw[0] = 1;
   for (int i=1; i=maxc && (ch-=maxc) )
            h[i][v] = ((LL)h[i][u]*seed + ch)%mod;
         DFS(v);
     }
}

inline int Min(int x, int y) {return x=0; i--) {
                if (Min(dep[u],dep[v]) < 1<\n");

       }

    }
    return 0;
}

C-平分游戏

正解讲的不太好，推荐一个很好的blog：点击打开链接

注： n == k 的情况应该直接输出gg吧，但加上这句我就gg了。。 去掉之后AC了。。

#include 
#include 
#include 
#include 
using namespace std;

typedef long long LL;
const int maxn = 1e6+5;
bool vis[maxn];
LL a[maxn];
LL sum,avg,ans;
int n,k;
vectorg;
bool flag;

LL cal(){
   int sz = g.size();
   g[sz-1] = 0;
   sort(g.begin(),g.end());
   return g[sz/2];
}

int main(){
    while (scanf("%d%d",&n,&k)==2) {
       // if (n == k) { printf("gg\n"); continue; }
        k = min(n-1,k);

        flag = 1;
        sum = 0; ans = 0;
        for (int i=0; i

D-psd面试

#include 
#include 
#include 
using namespace std;
 
const int maxn = 1500;
int dp[maxn][maxn];
char s[maxn];
int n;
 
int main(){
    while (scanf("%s",s)==1) {
        n = strlen(s);
        memset(dp,0,sizeof(dp));
        for (int i=0; i='A' && s[i]<='Z') s[i] = s[i]-'A'+'a';
 
        for (int i=0; i

E-回旋星空

#include 
#include 
using namespace std;
 
const int maxn = 1000+10;
struct Point{
    int x,y;
}p[maxn];
 
int n,ans;
int x,y;
int d[maxn];
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        for (int i=1; i<=n; i++) scanf("%d%d",&p[i].x,&p[i].y);
 
        ans = 0;
        for (int i=1; i<=n; i++) {
           x = p[i].x;
           y = p[i].y;
           for (int j=1; j<=n; j++) d[j] = (x-p[j].x)*(x-p[j].x) + (y-p[j].y)*(y-p[j].y);
           sort(d+1,d+n+1);
           int pt = 1;
           for (int j=2; j<=n; j++) if (d[pt]==d[j]) ans++; else pt = j;
        }
 
        if (ans==0) printf("WA\n"); else printf("%d\n",ans*2);
    }
    return 0;
}

F-等式

#include 
#include 
using namespace std;
 
const int maxn = 70;
int a[maxn],tot;
int ans,n;
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&n);
        tot = 0;
        memset(a,0,sizeof(a));
        for (int i=2; i*i<=n; i++) if (n%i==0) {
            while (n%i==0) a[tot]++, n/=i;
            tot++;
        }
        if (n>1) a[tot++]++;
 
        for (int i=0; i

G-旋转矩阵

#include 
#include 
#include 
#include 
using namespace std;
 
const int maxL = 10000;
const int maxn = 40;
char s[maxn][maxn];
int n,m;
char str[maxL];
stackst;
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d%d",&n,&m);
        for (int i=1; i<=n; i++) scanf("%s",s[i]+1);
        while (!st.empty()) st.pop();
        scanf("%s",str);
        int len = strlen(str);
        for (int i=0; i=1; i--)  {
                for (int j=m; j>=1; j--) printf("%c",s[i][j]);
                printf("\n");
            }
            printf("\n");
            continue;
        }
 
        printf("%d %d\n",m,n);
        if (st.top()==0 && sz==1 || st.top()==1 && sz==3) {
               for (int j=m; j>=1; j--) {
                  for (int i=1; i<=n; i++) printf("%c",s[i][j]);
                  printf("\n");
               }
               printf("\n");
               continue;
        }
 
        for (int j=1; j<=m; j++) {
            for (int i=n; i>=1; i--) printf("%c",s[i][j]);
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}

H-哲哲的疑惑

正解的数学变换非常复杂，我是看不懂了。。

推荐一个博客，讲述的不错，虽然思维跳跃大，但能直观的知道最后结果是怎么来的。

点击打开链接

#include 
#include 
using namespace std;

typedef long long LL;
LL n,m,l;
const LL p = 998244353;

LL quick_pow(LL a, LL b) {
   LL res = 1;
   while (b) {
      if (b&1) res = (res*a)%p;
      b>>=1;
      a = (a*a)%p;
   }
   return res;
}

LL C(LL n , LL m) {
     if (nn-m; i--) ans  = ans*i%p;
     return ans;
}

int main(){
    while (scanf("%lld%lld%lld",&n,&m,&l)==3) {
        printf("%lld\n",quick_pow(n-m,l)*C(n,m)%p);
    }
    return 0;
}

I-填空题

#include 
using namespace std;
 
int main(){
    printf("ac\n");
    return 0;
}

J-强迫症的序列

#include 
#include 
#include 
using namespace std;
const int maxn = 100000+5;
int a[maxn];
int n;
int total,t;
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        for (int i=1; i<=n; i++) scanf("%d",&a[i]);
        total = 0;
        sort(a+1,a+n+1);
        int sum = 0;
        t = a[n]-a[1];
        total += a[n]-a[1];
        for (int i=1; i

K-密码

#include 
#include 
#include 
using namespace std;
 
const int maxn = 100000+10;
int n,len,start;
char s[maxn];
vector G[maxn];
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&n);
        scanf("%s",s);
        len = strlen(s);
        for (int i=1; i<=n; i++) G[i].clear();
 
        if (n==1)  { printf("%s\n",s); continue; }
 
        bool dir = 1;
        int cnt = 1;
        for (int i=0; in) {cnt = n-1; dir=0;}
            }
            else if (cnt>=1 && dir==0) {
                G[cnt].push_back(s[i]);
                cnt--;
                if (cnt<1) {cnt = 2; dir = 1;}
            }
        }
        for (int i=1; i<=n; i++) {
            for (int j=0; j

L-用来作弊的药水

#include 
#include 
using namespace std;
 
const double eps = 1e-3;
int x,y,a,b;
double p,q;
 
int main(){
    int T;
    scanf("%d",&T);
    while (T--) {
        scanf("%d%d%d%d",&x,&a,&y,&b);
        p = (double)a*log2(x);
        q = (double)b*log2(y);
        if (fabs(p-q)

最后附上一张补完题目的图，这次比赛学到了不少东西吧，补完题也挺有成就感的。