“华为杯”大连理工大学第12届大学生程序设计大赛

• E 膜法项链
• J 更强的未来道具FG-C193
• K 取模运算

E 膜法项链

#include <cstdio>
#include <algorithm>
using namespace std;
#define Nmax 1000010
int A[Nmax*2];
int N, W;
long long cnt[Nmax * 2];
void adin(int x, int lx, int rx, int a) {
    int l, r;
    l = x - lx;
    r = max(rx - max(x, N+1), 0);
    cnt[rx-r-x+1] += a;
    cnt[rx-r-x+1+r] -= a;
    cnt[rx-r-x+1+l] -= a;
    cnt[rx-r-x+1+l+r] += a;
}
int main() {
    scanf("%d %d", &N, &W);
    for (int i = 1; i <= N; i++) {
        scanf("%d", A+i);
        A[N+i] = A[i];
    }
    int ls[Nmax][2], t = 0;
    ls[0][0] = 0;
    ls[0][1] = 0;
    t = 1;
    for (int i = 1; i <= N*2; i++) {
        while (ls[t-1][0] >= A[i]) {
            t--;
            adin(ls[t][1], ls[t-1][1], i, ls[t][0]);
        }
        ls[t][0] = A[i];
        ls[t][1] = i;
        t++;
    }
    while (t > 1) {
        t--;
        adin(ls[t][1], ls[t-1][1], N*2+1, ls[t][0]);
    }
    for (int i = 2; i <= N; i++)
        cnt[i] += cnt[i-1];
    for (int i = 2; i <= N; i++)
        cnt[i] += cnt[i-1];
    long long ans = 0x7fffffffffffffffll;
    for (int i = 1; i <= N; i++) {
        ans = min(ans, cnt[i] + (long long)W*(i-1));
    }
    printf("%lld\n", ans);
    return 0;
}

J 更强的未来道具FG-C193

#include <cstdio>
#include <algorithm>
using namespace std;
#define Nmax 100010
int N;
int re[Nmax*2];
int len2;
int fa[Nmax*2];
int cnt[Nmax*2];
int FA(int x) {
    if (fa[x] == x)
        return x;
    else
        return fa[x] = FA(fa[x]);
}
void marge(int x, int y) {
    x = FA(x);
    y = FA(y);
    if (x != y) {
        if (cnt[x] < cnt[y])
            fa[x] = y;
        else if (cnt[y] < cnt[x])
            fa[y] = x;
        else {
            fa[y] = x;
            cnt[x]++;
        }
    }
}
void manacher() {
    int r = -1, c = -1;
    int l;
    for (int i = 0; i < len2; i++) {
        if (i <= r) {
            l = min(re[c-(i-c)], r-i+1);
        }
        else
            l = 1;
        while (l < re[i]) {
            marge(i-l, i+l);
            l++;
        }
        if (i+re[i]-1 > r) {
            r = i+re[i]-1;
            c = i;
        }
    }
}
char s[Nmax*2];
int main() {
    scanf("%d", &N);
    len2 = N*2+1;
    for (int i = 0; i < len2; i++)
        fa[i] = i;
    re[0] = 1;
    for (int i = 1; i < len2; i++) {
        scanf("%d", re+i);
        if (i&1) {
            re[i] = re[i]*2;
        }
        else {
            re[i] = re[i]*2+1;
        }
    }
    manacher();
    int M;
    scanf("%d", &M);
    int x;
    char c;
    int ans = 0;
    for (int i = 0; i < M; i++) {
        scanf("%d %c", &x, &c);
        x = FA(1+x*2);
        if (s[x]) {
            if (s[x] != c)
                ans = -1;
        }
        else {
            s[x] = c;
        }
    }
    if (ans == -1) {
        printf("-1\n");
        return 0;
    }
    for (int i = 0; i < N; i++) {
        if (!s[FA(1+i*2)]) {
            printf("%d ", i);
            ans++;
        }
    }
    if (!ans) {
        for (int i = 0; i < N; i++) {
            printf("%c", s[FA(1+i*2)]);
        }
    }
    return 0;
}

K 取模运算

#include <cstdio>
using namespace std;
#define spl 317
int ls[100010];
int cnt[100010];
int mp[spl+3][spl+3];
int N, M;
int main() {
    scanf("%d %d", &N, &M);
    int a;
    for (int i = 1; i <= N; i++) {
        scanf("%d", &a);
        ls[i] = a; 
        cnt[a]++;
        for (int i = 1; i < spl; i++) {
            mp[i][a%i]++;   
        }
    }
    int op, x, y;
    int ans;
    for (int i = 0; i < M; i++) {
        scanf("%d %d %d", &op, &x, &y);
        if (op == 1) {
            cnt[ls[x]]--;
            for (int i = 1; i < spl; i++)
                mp[i][ls[x]%i]--;
            ls[x] = y;
            cnt[y]++;
            for (int i = 1; i < spl; i++) {
                mp[i][y%i]++;
            }
        }
        else {
            if (y < spl) {
                ans = mp[y][x];
            }
            else {
                ans = 0;
                for (; x <= 100000; x += y) {
                    ans += cnt[x];
                }
            }
            printf("%d\n", ans);
        }
    }
    return 0;
}