(div2)

1 2 1 2这样加就可以了

#include 
     
      typedef long long ll;const int N = 1e5 + 5;int main() {    int n; scanf ("%d", &n);    int ans = n / 3 * 2;    if (n % 3) {        ans++;    }    printf ("%d\n", ans);    return 0;}
     

 (div2)

水题，暴力模拟一下

#include 
     
      typedef long long ll;const int N = 1e5 + 5;char str[N];int a[N];int main() {    int n; scanf ("%d", &n);    scanf ("%s", str);    for (int i=0; i
      
       = n) {            break;        }        if (a[now] == -1) {            puts ("INFINITE");            return 0;        }        if (str[now] == '>') {            int pre = now;            now = now + a[now];            a[pre] = -1;        } else {            int pre = now;            now = now - a[now];            a[pre] = -1;        }    }    puts ("FINITE");    return 0;}
      
     

构造  (div2)

倒过来做，循环也反着来

#include 
     
      typedef long long ll;const int N = 1e2 + 5;const int Q = 1e4 + 5;int a[N][N];int t[Q], row[Q], col[Q], x[Q];int main() {    int n, m, q; scanf ("%d%d%d", &n, &m, &q);    for (int i=1; i<=q; ++i) {        scanf ("%d", t+i);        if (t[i] == 1) {            scanf ("%d", row+i);        }        if (t[i] == 2) {            scanf ("%d", col+i);        }        if (t[i] == 3) {            scanf ("%d%d%d", row+i, col+i, x+i);        }        //printf ("%d %d %d %d\n", t[i], row[i], col[i], x[i]);    }    for (int i=q; i>=1; --i) {        if (t[i] == 1) {            int last = a[row[i]][m];            for (int j=m; j>=2; --j) {                a[row[i]][j] = a[row[i]][j-1];            }            a[row[i]][1] = last;        }        if (t[i] == 2) {            int last = a[n][col[i]];            for (int j=n; j>=2; --j) {                a[j][col[i]] = a[j-1][col[i]];            }            a[1][col[i]] = last;        }        if (t[i] == 3) {            a[row[i]][col[i]] = x[i];        }    }    for (int i=1; i<=n; ++i) {        for (int j=1; j<=m; ++j) {            printf ("%d%c", a[i][j], j == m ? '\n' : ' ');        }    }    return 0;}
     

数学  (div2)

题意：男生与女生围成圈跳舞，女生的位置不变，男生可以移动x个女生或者相邻的男生奇偶互换，问最后男生的排列

分析：问题的关键点在于奇数男生的圈顺序不变，偶数也不变，只是起点的位置改变，所以只要对两个起点操作就行了。

#include 
     
      typedef long long ll;const int N = 1e6 + 5;int ans[N];int main() {    int p0 = 0, p1 = 1;    int n, q; scanf ("%d%d", &n, &q);    for (int i=0; i
     

数学+前（后）缀  (div1)

题意：已知p(max(a,b)=k) 和 p(min(a,b)=k)的概率，求p(a=k) 和 p(b=k)

分析：

P(a = k) = P(a <= k) — P(a <= k-1) P(max(a, b) <= k) = P(a <= k) * P(b <= k)

P(min(a, b) >= k) = P(a >= k) * P(b >= k) = (1 — P(a <= k-1)) *(1 — P(b <= k-1))

即

　

解方程的和，从而求得和

#include 
     
      const int N = 1e5 + 5;double p[N], q[N], a[N], b[N];int main() {    int n; scanf ("%d", &n);    for (int i=1; i<=n; ++i) {        scanf ("%lf", p+i);        p[i] += p[i-1];    }    for (int i=1; i<=n; ++i) {        scanf ("%lf", q+i);    }    for (int i=n; i>=1; --i) {        q[i] += q[i+1];    }    for (int i=1; i<=n; ++i) {        double A = p[i], B = q[i+1];        double C = B - A - 1;        double delta = sqrt (std::max (C*C - 4 * A, 0.0));        a[i] = (-C+delta) / 2;        b[i] = (-C-delta) / 2;    }    for (int i=1; i<=n; ++i) {        printf ("%.10f%c", a[i] - a[i-1], i == n ? '\n' : ' ');    }    for (int i=1; i<=n; ++i) {        printf ("%.10f%c", b[i] - b[i-1], i == n ? '\n' : ' ');    }    return 0;}