Description

BZOJ1044

Luogu2511

Solution

第一问二分答案，第二问就是一个简单的DP，前缀和优化一下就行。

Code

const int N = 5e4 + 10;
const int ha = 1e4 + 7;

int a[N], f[N], s[N], sf[N], nxt[N];
int n, m;

bool check(int x) {
    int l = 1, r = 0, ans = 0;
    while (r <= n) {
        while (r <= n && s[r] - s[l - 1] <= x) r++;
        // if (x == 1) printf("%d %d\n", l, r);
        ans++;
        if (r > n) break;
        if (l == r) return false;
        l = r;
    }
    return ans <= m;
}
int solve(int x) {
    f[1] = 1;
    for (int i = 1; i <= n; ++i) sf[i] = 1;
    int ans = 0;
    int k = n - 1;
    for (int j = n; j > 1; --j) {
        while (k > 0 && s[j] - s[k] <= x) k--;
        nxt[j] = k;
    }
    for (int i = 1; i <= m; ++i) {
        for (int j = n; j > 1; --j) {
            f[j] = 0;
            f[j] = (sf[j - 1] - sf[nxt[j]] + ha);
            if (f[j] >= ha) f[j] -= ha;
        }
        f[1] = sf[1] = 0;
        for (int j = 2; j <= n; ++j) {
            sf[j] = (sf[j - 1] + f[j]);
            if (sf[j] >= ha) sf[j] -= ha;
        }
        (ans += f[n]) %= ha;
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    m++;
    n++;
    for (int i = 2; i <= n; ++i) {
        scanf("%d", &a[i]);
        s[i] = s[i - 1] + a[i];
    }
    int l = 0, r = s[n], ans = -1;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) {
            ans = mid;
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    printf("%d %d\n", ans, solve(ans));
    return 0;
}

[HAOI2008]木棍分割

原文：https://www.cnblogs.com/wyxwyx/p/bzoj1044.html