codeforces 733E (数学)

题目链接：点击这里

题意：ｎ个台阶，每个台阶上有上或者下，每经过一次都会改变方向。求从每个台阶出发，从最下方或者最上方走出去的步数

对于每一个开始的台阶，只有他下方的Ｕ和他上方的Ｄ会影响最后的步数。如果要从下方出去就需要把下面都变成Ｄ，同样的，要从上面出去就要把上面都变成Ｕ。观察一下发现交换初始台阶上下方的Ｄ和Ｕ的步数花费就是两者距离的一半，最后加上开始台阶从下方或者上方走出的步数。

trick:结果会爆int

#include <bits/stdc++.h>
#define Clear(x,y) memset (x,y,sizeof(x))
#define FOR(a,b,c) for (int a = b; a <= c; a++)
#define REP(a,b,c) for (int a = b; a >= c; a--)
#define first fi
#define second se
#define pii pair<int, int>
#define pli pair<long long, int>
#define mp make_pair
#define pb push_back
using namespace std;
#define maxn 1000005

char s[maxn];
long long n, sum1, sum2;
long long l[maxn], r[maxn];
vector <long long> ans1, ans2, u, d;

int main () {
//freopen ("more.in", "r", stdin);
cin >> n >> s+1;
l[0] = r[n+1] = 0;
u.clear (); d.clear ();
FOR (i, 1, n) {
l[i] = l[i-1]+(s[i] == 'U');
if (s[i] == 'U') u.pb (i);
else d.pb (i);
}
REP (i, n, 1) {
r[i] = r[i+1]+(s[i] == 'D');
}
int cut = n+1;
FOR (i, 1, n) {
if (l[i-1] < r[i+1] || (l[i-1] == r[i+1] && s[i] == 'D')) continue;
else {
cut = i;
break;
}
}
sum1 = sum2 = 0;
ans1.clear (); ans2.clear ();
int i = u.size()-1, j = 0;
FOR (pos, 1, cut-1) {
if (s[pos] == 'U') sum1 += pos*2;
else sum2 -= pos*2;
sum2 += d[j++]*2;
ans1.pb (sum2-sum1+pos);
}
sum1 = sum2 = 0;
REP (pos, n, cut) {
if (s[pos] == 'D') sum2 += pos*2;
else if (s[pos] == 'U') sum1 -= pos*2;
sum1 += u[i--]*2;
ans2.pb (sum2-sum1+n-pos+1);
}
for (int i = 0; i < ans1.size (); i++) cout << ans1[i] << " ";
REP (i, ans2.size()-1, 0) {
cout << ans2[i] << " ";
} cout << endl;
return 0;
}