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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#215352#2777. 2048KXG00ms0kbC++113.5kb2024-11-28 20:31:262024-11-28 23:11:58

answer

#include <cstdio>
#include <cstring>
#include <vector>
#include <cmath>
using namespace std;
int n, p, a[10010], s[10010];
int pos[10010];
int dp[35000010];
unsigned highbit(unsigned i) {
    i |= i >> 1;
    i |= i >> 2;
    i |= i >> 4;
    i |= i >> 8;
    i |= i >> 16;
    i |= i >> 32;
    return (i + 1) >> 1;
}
unsigned lowbit(unsigned i) {
    return i & (-i);
}
int getnxt(int S, int x) {
    if (S != 0 && lowbit(S) < x) {
        return -1;
    }
    S += x;
    if (S >= (1 << p)) return -2;
    return S;
}
int main() {
    memset(pos, -1, sizeof(pos));
    scanf("%d", &n);
    pos[0] = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        s[i] = s[i - 1] + a[i];
        pos[s[i]] = i;
    }
    p = log2(s[n]);
    if (s[n] != (1 << p)) {
        printf("no\n");
        return 0;
    }
    if (n == 1) {
        printf("l\n");
        return 0;
    }
    for (int S = 0; S < (1 << (2 * p)); S++) {
        dp[S] = -1;
    }
    dp[0] = 0;
    int anslst = -1;
    for (int S1 = 0; S1 < (1 << p); S1++) {
        for (int S2 = 0; S1 + S2 < (1 << p) && S2 <= (1 << (p - 1)); S2++) {
            if (pos[S1 + S2] == -1 || (highbit(S1) == highbit(S2) && S1 != 0 && S2 != 0)) {
                continue;
            }
            int S = (S1 << (p - 1)) | S2;
            if (dp[S] == -1) continue;
            // printf("%d %d : %d\n", S1, S2, dp[S]);
            int ps = pos[S1 + S2];
            int nxtS1, nxtS2;
            nxtS1 = getnxt(S1, a[ps + 1]);
            nxtS2 = S2;
            // printf("%d %d : %d %d\n", S1, S2, nxtS1, nxtS2);
            if (nxtS1 != -1) {
                if (nxtS1 == -2) {
                    anslst = (S << 1);
                } else {
                    if (highbit(nxtS1) == highbit(nxtS2)) {
                        int x = highbit(nxtS1);
                        nxtS1 ^= x;
                        nxtS2 ^= x;
                        nxtS1 ^= (x << 1);
                    }
                    if (nxtS1 >= (1 << p) || nxtS2 >= (1 << p)) {
                        anslst = (S << 1);
                    }
                    dp[(nxtS1 << (p - 1)) | nxtS2] = (S << 1);
                }
            }
            nxtS1 = S1;
            nxtS2 = getnxt(S2, a[ps + 1]);
            // printf("%d %d : %d %d\n", S1, S2, nxtS1, nxtS2);
            if (nxtS2 != -1) {
                if (nxtS2 == -2) {
                    anslst = (S << 1) | 1;
                } else {
                    if (highbit(nxtS1) == highbit(nxtS2)) {
                        int x = highbit(nxtS1);
                        nxtS1 ^= x;
                        nxtS2 ^= x;
                        nxtS1 ^= (x << 1);
                    }
                    if (nxtS1 >= (1 << p) || nxtS2 >= (1 << p)) {
                        anslst = (S << 1) | 1;
                    }
                    dp[(nxtS1 << (p - 1)) | nxtS2] = (S << 1) | 1;
                }
            }
        }
    }
    if (anslst == -1) {
        printf("no\n");
    } else {
        vector<int> ans;
        int now = anslst;
        while (true) {
            ans.push_back(now & 1);
            now >>= 1;
            if (now == 0) {
                break;
            }
            now = dp[now];
        }
        for (int i = ans.size() - 1; i >= 0; i--) {
            if (ans[i] == 0) {
                printf("l");
            } else {
                printf("r");
            }
        }
    }
    return 0;
}

详细

小提示:点击横条可展开更详细的信息

Subtask #1:

score: 0
Runtime Error

Test #1:

score: 0
Runtime Error

input:

20
2 8 8 256 2 32 64 64 2 2 64 8 256 128 128 1024 2048 2048 1024 1024

output:


result:


Subtask #2:

score: 0
Runtime Error

Test #11:

score: 0
Runtime Error

input:

50
1 1 2 2 8 8 8 8 1 1 16 8 8 4 4 8 8 8 64 32 8 8 16 256 256 64 64 128 4 4 256 128 128 256 256 1024 ...

output:


result:


Subtask #3:

score: 0
Runtime Error

Test #21:

score: 0
Runtime Error

input:

1000
1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 2 2 1 1 1 1 8 8 16 1 1 1 1 1 1 1 1 16 8 1 1 1 1 1 1 2 8 8 4 4 4 4...

output:


result: