2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)

A. Find a Number

找到一个树，可以被d整除，且数字和为s

记忆化搜索

     static class S{
         int mod,s;
         String str;

         public S(int mod, int s, String str) {
             this.mod = mod;
             this.s = s;
             this.str = str;
         }
     }

     public static void main(String[] args) {
         IO io = new IO();
         int[][]vis=new int[550][5500];
         int d=io.nextInt(),s=io.nextInt();
         Queue<S>q=new ArrayDeque<>(10000);
         q.add(new S(0,0,""));
         while (!q.isEmpty()){
             S cur=q.poll();
             if (cur.mod==0&&cur.s==s){
                 io.println(cur.str);return;
             }
             for (int i = 0; i <=9; i++) {
                 int mm=(cur.mod*10+i)%d;
                 int ss=cur.s+i;
                 if (vis[mm][ss]==0&&ss<=s){
                     q.add(new S(mm,ss,cur.str+i));
                     vis[mm][ss]=1;
                 }
             }
         }
         io.println(-1);
     }

B. Berkomnadzor——我选择狗带……这题目有毒啊

C. Cloud Computing

有m个计划，每个计划的内容是从[l，r]天内，总共有c个处理器，每个p元。问，从[1，n]天，每天买k个处理器（尽量买齐k个）最小花费是多少

线段树表示当天有效的计划，处理器价格就是叶子编号。注意此题要全部开long，tre[i][0]是个数总和，tre[i][1]是价值总和。（才发现把tre拆成两个数组会快一倍）

     private static final int c = 1000100;
     static long[] S = new long[c << 2];
     static long[] sum = new long[c << 2];
     static long ans;
     static ArrayList<long[]>[] plan = new ArrayList[c];

     static void update(int t, int l, int r, long c, long p) {
         S[t] += c;
         sum[t] += c * p;
         if (l == r) return;
         int mid = l + r >> 1;
         if (p <= mid) update(t << 1, l, mid, c, p);
         else update(t << 1 | 1, mid + 1, r, c, p);
     }

     static long query(int t, int l, int r, long k) {
         if (k <= 0) return 0;
         if (S[t] <= k) return sum[t];
         if (l == r) return k * l;
         int mid = l + r >> 1;
         return query(t << 1, l, mid, k) +
                 query(t << 1 | 1, mid + 1, r, k - S[t << 1]);
     }

     public static void main(String[] args) {
         for (int i = 0; i < plan.length; i++) plan[i] = new ArrayList<>();
         IO io = new IO();
         int n = io.nextInt(), k = io.nextInt(), m = io.nextInt();
         for (int i = 0; i < m; i++) {
             int l = io.nextInt(), r = io.nextInt(), c = io.nextInt(), p = io.nextInt();
             plan[l].add(new long[]{c, p});
             plan[r + 1].add(new long[]{-c, p});
         }
         for (int i = 1; i <= n; i++) {
             for (long[] v : plan[i]) update(1, 1, c, v[0], v[1]);
             ans += query(1, 1, c, Math.min(S[1], k));
         }
         io.println(ans);
     }

D. Garbage Disposal

有n天，每天产生ni的垃圾，每个垃圾袋容量为k，每天产生的垃圾最迟第二天要丢掉，第n天的垃圾当天要丢掉，问使用的垃圾袋最少个数

模拟

     private static final int c = (int) 2e6;

     public static void main(String[] args) {
         IO io=new IO();
         long n=io.nextLong(),k=io.nextLong();
         long[]a=new long[2];
         long ans=0;
         for (int i=1;i<=n;i++){
             int now=i&1,pre=now^1;
             long t=io.nextLong();
             if (i==n){
                 ans+=Math.ceil((double)(t+a[pre])/k);
                 break;
             }
             if (a[pre]!=0){
                 ans+=Math.ceil((double) a[pre]/k);
                 long more=k-a[pre]%k;
                 t-=more;
                 if (t<0)t=0;
             }
             ans+=t/k;
             t%=k;
             a[now]=t;
         }
         io.println(ans);
     }