Selection (opentrains)

这个本来是一道水题，唯一的坑就是shift只能操作一次
然后用f[i]记录前i个字符中 ‘*’ 的数量，g[i]记录前i个字符中 ‘.’ 的数量(当然也可以直接用i-f[i]运算)
假设当前位置为i，要[x,i]作为shift的区段，操作数为f[n]-f[i]+f[x-1]+g[i]-g[x-1]+2，可以转化一下成：f[n]-f[i]+g[i]+f[x-1]-g[x-1]+2，其中i为枚举变量，只要维护 f[x-1]-g[x-1] 尽量大就好了，最终是O(n)的

#include <bits/stdc++.h>#include <string.h>using namespace std;const int maxn = 100010;char s[maxn];int n, f[maxn], ans, t, l, x, y, g[maxn], sum;int main() {    freopen("input.txt","r",stdin);    freopen("output.txt","w",stdout);    scanf("%d", &n);    scanf("%s", s+1);    l = 1;    x = 0; y = 0;    for (int i = 1; i <= n; i++) {        f[i] = f[i-1];        g[i] = g[i-1];        if (s[i] == '*') f[i]++;        else g[i]++;    }    ans = f[n];    for (int i = 1; i <= n; i++) {        if (i > 1) {            //l, i            if (f[n]-f[i]+g[i]-g[l-1]+f[l-1]+2 < ans) {                ans = f[n]-f[i]+g[i]-g[l-1]+f[l-1]+2;                x = l;                y = i;            }        }        if (f[i-1]-g[i-1] < f[l-1]-g[l-1]) l = i;    }    sum = 0;    for (int i = 1; i < x; i++) if (s[i] == '*') sum++;    for (int i = x; i <= y; i++) if (s[i] == '.') sum++;    for (int i = y+1; i <= n; i++) if (s[i] == '*') sum++;    if (x != 0) sum += 2;    if (sum != ans) while (1) {}    printf("%d\n", ans);    bool flag = false;    if (x != 0) {        printf("%d\n", x);        printf("Shift+%d\n", y);        flag = true;    }    for (int i = 1; i < x; i++) if (s[i] == '*') {        if (flag) printf("Ctrl+%d\n", i);        else printf("%d\n", i);        flag = true;    }    for (int i = x; i <= y; i++) if (s[i] == '.') {        if (flag) printf("Ctrl+%d\n", i);        else printf("%d\n", i);        flag = true;    }    for (int i = y+1; i <= n; i++) if (s[i] == '*') {        if (flag) printf("Ctrl+%d\n", i);        else printf("%d\n", i);        flag = true;    }}