Codeforces #418(Div 2)

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C. An impassioned circulation of affection

预处理n|Σ|种询问，每次用双指针查找。

#include <bits/stdc++.h>using namespace std;const int MAXN = 1505;int n, q;char str[MAXN];int A[MAXN][26];int pos[MAXN], top = 0;void init(){    for (int i = 1; i <= n; i++) {        for (int j = 0; j < 26; j++) {            top = 0;            memset(pos, 0, sizeof pos);            for (register int k = 1; k <= n; k++)                if (str[k]-'a' == j)                    pos[k] = 1;            int l = 1, len = 0, t = 0, ans = 0;            for (register int k = 1; k <= n; k++) {                len += 1-pos[k], t++;                while (len > i) len -= 1-pos[l], t--, l++;                ans = max(ans, t);            }            A[i][j] = ans;        }    }}int main(){    scanf("%d", &n);    scanf("%s", str+1);    init();    scanf("%d", &q);    for (int i = 1; i <= q; i++) {        int u; char c;        scanf("%d %c", &u, &c);        printf("%d\n", A[u][c-'a']);    }    return 0;}

D. An overnight dance in discotheque

大力贪心，不会证明。

#include <bits/stdc++.h>using namespace std;const int MAXN = 1005;const double eps = 1e-9;struct circ {    int x, y, r;    friend bool operator < (const circ &a, const circ &b)    { return a.r > b.r; }} c[MAXN], A[MAXN], B[MAXN];int n;int ta = 0, tb = 0;int main(){    scanf("%d", &n);    for (int i = 1; i <= n; i++)         scanf("%d%d%d", &c[i].x, &c[i].y, &c[i].r);    sort(c+1, c+n+1);    long long ans = 0;    for (int i = 1; i <= n; i++) {        int tmsa = 0, tmsb = 0;        for (int j = 1; j <= ta; j++)            if ((long long)(A[j].x-c[i].x)*(A[j].x-c[i].x)+(long long)(A[j].y-c[i].y)*(A[j].y-c[i].y) <= (long long)(A[j].r-c[i].r)*(A[j].r-c[i].r)) tmsa++;        for (int j = 1; j <= tb; j++)            if ((long long)(B[j].x-c[i].x)*(B[j].x-c[i].x)+(long long)(B[j].y-c[i].y)*(B[j].y-c[i].y) <= (long long)(B[j].r-c[i].r)*(B[j].r-c[i].r)) tmsb++;        if ((tmsa&1)==0) A[++ta] = c[i], ans += (long long)c[i].r*c[i].r;        else if ((tmsb&1)==0) B[++tb] = c[i], ans += (long long)c[i].r*c[i].r;        else A[++ta] = c[i], ans -= (long long)c[i].r*c[i].r;    }    printf("%.10f\n", ans*M_PI);    return 0;}