Astar Round2B

来源：互联网发布：plc编程实例编辑：程序博客网时间：2024/05/16 07:22

放在前面..

傻逼呵呵还继续做百度之星..

比赛感悟啦啦啦..

因为上一场傻逼呵呵rank610.. 所以这一场还来.. ~~说实话还是很想要T恤哒..~~

今天又和HWizard大神一起打啦啦啦.. 照样的naive啊..

这次服务器没有卡爆啊..

这一次全部题都先看一遍.. 以免像上一场一样最水的1003还没看就炸了..

一看，这场1003好像也挺水的.. 找规律嘛.. 然后就进入乱来找规律的time了.. 找着找着，就听到了HWizard来了一句，杨辉三角！然后一看我的规律表，对啊杨辉三角..

10min后第一次提交.. WA了.. 检查代码.. 我擦算错了，改啊改.. 改对了..

又浏览一次题目.. 哎1006不是原题吗.. 好像某次在某牛做过诶.. 只是数据大了点5000.. 码代码，自信提交.. 我擦这数据怎么改8000了啊.. 啊改改改.. AC了..

继续看题.. 群上说1005是普及水平题.. ahhh来算算.. 哎贪心一下好像行的样子.. 继续码代码..

10min后提交.. WA.. 改long long再交.. 还WA.. 卧槽这什么鬼.. 这个做法绝对没错啊.. 哎不对.. 这个freopen是smg.. AC了.. 成功罚了2次时..

看了一下排名.. rank25.. 好开森..

剩下1h了.. 看下别的题.. 没思路.. ~~教一下wohenshuai1005~~

啊啊啊别的题都不会啊.. 算了死磕1001..

参考了下省队队长的想法.. 好像有道理.. 开始码代码..

15min后打完.. 就要结束了啊！！交交交！！

还是超时了.. 没交成功.. 后面在hdu官网交了.. AC了.. 就差那么点啊！！

最后rank90.. 还算好吧.. 起码比上一场好.. ~~rating又涨了.. 成功紫名.. rp--~~

题解一发..

1001

省队队长的题解：

我觉得讲的很清晰.. 补充一下就是找第一个比它大（小）的时候就是把这个当做最大（小）值..

code

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#define LL long longusing namespace std;const LL Maxn = 110000;LL mx[Maxn], mn[Maxn];LL n;LL a[Maxn];LL ans[Maxn];LL _min ( LL x, LL y ){ return x < y ? x : y; }LL _max ( LL x, LL y ){ return x > y ? x : y; }int main (){    LL i, j, k;    while ( scanf ( "%I64d", &n ) != EOF ){        for ( i = 1; i <= n; i ++ ) scanf ( "%I64d", &a[i] );        for ( i = 1; i <= n; i ++ ){            mx[i] = mn[i] = n+1;            for ( j = i+1; j <= n; j ++ ){                if ( a[j] < a[i] ){                    mn[i] = j;                    break;                }            }            for ( j = i+1; j <= n; j ++ ){                if ( a[j] > a[i] ){                    mx[i] = j;                    break;                }            }        }        memset ( ans, 0, sizeof (ans) );        for ( i = 1; i <= n; i ++ ){            LL mmx = i, mmn = i;            while ( mx[mmx] != n+1 || mn[mmn] != n+1 ){                LL tmp = _min ( mx[mmx], mn[mmn] );                ans[tmp-i] = _max ( ans[tmp-i], a[mmx]*a[mmn] );                if ( mx[mmx] == tmp ) mmx = mx[mmx];                else mmn = mn[mmn];            }            ans[n-i+1] = _max ( ans[n-i+1], a[mmx]*a[mmn] );        }        for ( i = n; i >= 1; i -- ) ans[i] = _max ( ans[i], ans[i+1] );        for ( i = 1; i <= n; i ++ ) printf ( "%I64d\n", ans[i] );    }    return 0;}

1002

暂时还不会..

1003

推一下dp数组..

1 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 2 3 4 5 6 0 1 3 6 10 15 21 0 1 4 10 20 35 56 0 1 5 15 35 70 126 0 1 6 21 56 126 252

可以看出f[i][j]=f[i-1][j]+f[i][j-1].. 很符合杨辉三角的特性啊..

那么答案就是(n+m−4n−2)，来个逆元就好啦..

code

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#define LL long longusing namespace std;const LL Mod = 1000000007;LL n, m;LL A, B, X, Y;LL exgcd ( LL a, LL b, LL &x, LL &y ){    if ( a == 0 ){        x = 0; y = 1;        return b;    }    LL fx, fy;    LL d = exgcd ( b%a, a, fx, fy );    y = fx;    x = fy-(b/a)*fx;    return d;}LL ant ( LL x ){    A = x;    B = Mod;    LL d = exgcd ( A, B, X, Y );    return (X%B+B)%B;}LL C ( LL x, LL y ){    LL fz = 1;    LL i;    for ( i = y; i >= y-x+1; i -- ) fz = (fz*i)%Mod;    for ( i = 1; i <= x; i ++ ) fz = (fz*ant(i))%Mod;    return fz;}int main (){    LL i, j, k;    while ( scanf ( "%I64d%I64d", &n, &m ) != EOF ){        n --; m --;        printf ( "%I64d\n", C ( n-1, n+m-2 ) );    }    return 0;}

1004

暂时还不会做..

1005

首先我们先把这些区间按照左端点排个序，那么我们就可以很容易算出某k段区间的左端点了，那么只要在这些区间中找一个最左的右端点作为这k段区间的右端点.. 那么当当前存的区间数大于k时，就把右端点最靠左的给扔出去.. 这东西用堆维护就好啦..

code

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>#include <queue>#define LL long longusing namespace std;const LL Maxn = 110000;struct node {    LL l, r;}list[Maxn];LL a[Maxn], s[Maxn];LL n, K, m;bool cmp ( node x, node y ){    if ( x.l != y.l ) return x.l < y.l;    return x.r < y.r;}priority_queue < LL, vector<LL>, greater<LL> > q;LL _max ( LL x, LL y ){ return x > y ? x : y; }int main (){    LL i, j, k;    while ( scanf ( "%I64d%I64d%I64d", &n, &K, &m ) != EOF ){        s[0] = 0;        for ( i = 1; i <= n; i ++ ){            scanf ( "%I64d", &a[i] );            s[i] = s[i-1]+a[i];        }        for ( i = 1; i <= m; i ++ ) scanf ( "%I64d%I64d", &list[i].l, &list[i].r );        sort ( list+1, list+m+1, cmp );        while ( !q.empty () ) q.pop ();        LL ans = 0;        for ( i = 1; i <= m; i ++ ){            q.push (list[i].r);            if ( i > K ) q.pop ();            if ( i >= K ){                LL r = q.top ();                ans = _max ( ans, s[r]-s[list[i].l-1] );            }        }        printf ( "%I64d\n", ans );    }    return 0;}

1006

对于当前某一个数a[i]，我们把整个数列比它大的标为1，比它小的标为-1，那么符合条件的区间即为区间和为0的..

那么就根据这个就算咯.. 啦啦啦..

code

#include <cstdio>#include <cstring>#include <cstdlib>#include <algorithm>using namespace std;const int Maxn = 17000;int sum[Maxn], a[Maxn];int n, ans[Maxn];int main (){    int i, j, k;    while ( scanf ( "%d", &n ) != EOF ){        for ( i = 1; i <= n; i ++ ){            scanf ( "%d", &a[i] );        }        for ( i = 1; i <= n; i ++ ){            memset ( sum, 0, sizeof (sum) );            int s = 0;            for ( j = i; j <= n; j ++ ){                if ( a[j] > a[i] ) s ++;                else if ( a[j] < a[i] ) s --;                sum[s+8000] ++;            }            ans[i] = 0;            s = 0;            for ( j = i; j >= 1; j -- ){                if ( a[j] > a[i] ) s ++;                else if ( a[j] < a[i] ) s --;                ans[i] += sum[8000-s];            }        }        for ( i = 1; i < n; i ++ ) printf ( "%d ", ans[i] );        printf ( "%d\n", ans[n] );    }    return 0;}

最后再说几句..

那么这场比赛还是挺好的.. 但还是有很多地方自己做的不好（比如说对随机数据的处理）.. 还需要继续做多点题多积累点经验啊..

啦啦啦.. ~~rank90还是很开森的..~~

0 0