HAOI 2008 木棍分割

//需要优化很多的DP，包括预处理前缀和，滚动数组，预处理可转移状态 #include<bits/stdc++.h>using namespace std;const int moder = 10007;const int maxn = 50010;const int inf = 0x3f3f3f3f;int n, m, tmp;int f[maxn], a[maxn], last[maxn], tot[maxn], sum[maxn];inline bool check(int mid){    int tot = 0, final = 0;    for(int i = 1; i <= n; i++){        if(tot + a[i] <= mid)    tot += a[i];        else if(a[i] <= mid)    tot = a[i], final++;        else if(a[i] > mid)    return 0;        if(final > m)    return 0;    }    if(final == m)    return 1;}int main(){    scanf("%d%d", &n, &m);    int mina = inf;    for(int i = 1; i <= n; i++){        scanf("%d", &a[i]);        mina = min(a[i], mina);        sum[i] = sum[i-1] + a[i];    }    int ans;    int l = mina, r = sum[n] + 1;    while(l < r){        int mid = l + r >> 1;        if(check(mid)){            ans = mid;            r = mid;        }        else l = mid + 1;    }//第一问二分     for(int i = 1; i <= n; i++){        if(sum[i] <= ans)    f[i] = 1;//预处理ans从开始能到的状态         else break;    }    for(int i = 1; i <= n; i++)        tot[i] = (tot[i-1] + f[i]) % moder;//预处理前缀和     int now = 0;    for(int i = 1; i <= n; i++){        while(1){            if(sum[i] - sum[now] <= ans)    break;            else now++;        }        last[i] = now-1;//找每个i符合条件的转移状态     }    for(int i = 1; i <= m; i++){        for(int j = 1; j <= n; j++)            f[j] = (tot[j-1] - tot[last[j]] + moder) % moder;        for(int j = 1; j <= n; j++)            tot[j] = tot[j-1] + f[j];        tmp = (tmp + f[n]) % moder;//至多m次，所以每次都转移     }    printf("%d %d\n", ans, tmp);    return 0;}