Educational Codeforces Round 15

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A
题意：最长的连续LIS。
dp 跑一发即可

#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#include <string>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) {x += y; x %= MOD;}int dp[MAXN], a[MAXN];int main(){    int n;    while(scanf("%d", &n) != EOF) {        int ans = 0; dp[0] = 0;        for(int i = 1; i <= n; i++) {            scanf("%d", &a[i]);            if(i == 1 || a[i] > a[i-1]) {                dp[i] = dp[i-1] + 1;            }            else {                dp[i] = 1;            }            ans = max(ans, dp[i]);        }        printf("%d\n", ans);    }    return 0;}

B
题意：问你有多少对a[i]，a[j]使得二者和是2的幂。

#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#include <string>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) {x += y; x %= MOD;}map<LL, int> num;LL f[35];LL a[MAXN];int main(){    f[0] = 1LL;    for(int i = 1; i <= 31; i++) {        f[i] = f[i-1] * 2;    }    int n;    while(scanf("%d", &n) != EOF) {        num.clear();        for(int i = 1; i <= n; i++) {            scanf("%lld", &a[i]);            num[a[i]]++;        }        LL ans = 0;        for(int i = 1; i <= n; i++) {            for(int j = 1; j <= 31; j++) {                if(a[i] + a[i] == f[j]) {                    ans += num[a[i]] - 1;                }                else {                    ans += num[f[j] - a[i]];                }            }        }        printf("%lld\n", ans / 2);    }    return 0;}

C
题意：n个城市，m个塔在一条直线上。问塔最小的覆盖半径使得所有城市都被覆盖。
二分即可，比较坑的是，二分会爆int。。。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#include <string>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) {x += y; x %= MOD;}int x1[MAXN], x2[MAXN];int n, m;bool judge(int o) {    int i = 1, j;    for(j = 1; j <= m; j++) {        while(i <= n && abs(x1[i] - x2[j]) <= o) {            i++;        }    }    return i > n;}int main(){    while(scanf("%d%d", &n, &m) != EOF) {        int l1 = 1e9 + 10, r1 = -1e9 - 10;        for(int i = 1; i <= n; i++) {            scanf("%d", &x1[i]);            l1 = min(l1, x1[i]);            r1 = max(r1, x1[i]);        }        int l2 = 1e9 + 10, r2 = -1e9 - 10;        for(int i = 1; i <= m; i++) {            scanf("%d", &x2[i]);            l2 = min(l2, x2[i]);            r2 = max(r2, x2[i]);        }        LL l = 0, r = max(abs(r1 - l2), abs(r2 - l1)), ans;        while(r >= l) {            LL mid = (l + r) >> 1;            if(judge(mid)) {                ans = mid;                r = mid - 1;            }            else {                l = mid + 1;            }        }        printf("%lld\n", ans);    }    return 0;}

D
题意：一个人要走d千米的路，开车每千米花费a秒，步行每千米花费b秒。开车每次最多走k千米，而且走完k千米后必须花费t秒修车才可以继续开车，他可以在任意时刻丢下车步行（之后不能继续乘车）。问你最少需要多少时间到达目的地。

思路：特判d <= k。
走k千米开车最优时就开车，最后一段看是否等于k，不等于的话，取步行和开车（少一次修车时间）的最优值。走k千米步行最优时，就先开车k千米，然后开始步行。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#include <string>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) {x += y; x %= MOD;}int main(){    LL d, k, a, b, t;    while(scanf("%lld%lld%lld%lld%lld", &d, &k, &a, &b, &t) != EOF) {        if(d <= k) {            printf("%lld\n", d * a);            continue;        }        LL kk = k * (a - b) + t;        if(kk > 0) {            LL ans = kk + d * b - t;            printf("%lld\n", ans);        }        else {            LL ans = d / k * k * a + (d / k - 1) * t;            LL yu = d - d / k * k;            ans += min(t + yu * a, yu * b);            printf("%lld\n", ans);        }    }    return 0;}

题意：给你一个有向图以及边权，可能存在自环。问每个点的s和m信息
s信息：从当前点出发的长度为k的路径上的边权和。
m信息：同上，只不过找的是路径最小边权。

题目没有说长度为k的路径一定存在，出题人的意思应该是一定存在的。
思路：类似于倍增。
g[i][j]表示从i点出发走2j步到达的点。
sum[i][j]记录i−>g[i][j]路径权值和
Min[i][j]记录i−>g[i][j]路径最小边权。
然后统计所有的信息。
查询k时，枚举二进制位即可。

#include <iostream>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#include <cstring>#include <queue>#include <map>#include <set>#include <string>#define CLR(a, b) memset(a, (b), sizeof(a))#define ll o<<1#define rr o<<1|1using namespace std;typedef long long LL;typedef pair<int, int> pii;const int MAXN = 1e5 + 10;const int MAXM = 1e5 + 1;const int INF = 1e9 + 10;const int MOD = 1e9 + 7;void add(LL &x, LL y) {x += y; x %= MOD;}LL sum[MAXN][36];LL Min[MAXN][36];int g[MAXN][36];LL f[36];int main(){    f[0] = 1LL;    for(int i = 1; i <= 35; i++) {        f[i] = f[i-1] * 2;    }    int n; LL k;    while(scanf("%d%lld", &n, &k) != EOF) {        for(int i = 0; i < n; i++) {            scanf("%d", &g[i][0]);        }        for(int i = 0; i < n; i++) {            scanf("%lld", &Min[i][0]);            sum[i][0] = Min[i][0];        }        for(int j = 1; j <= 35; j++) {            for(int i = 0; i < n; i++) {                g[i][j] = g[g[i][j-1]][j-1];                sum[i][j] = sum[i][j-1] + sum[g[i][j-1]][j-1];                Min[i][j] = min(Min[i][j-1], Min[g[i][j-1]][j-1]);            }        }        for(int i = 0; i < n; i++) {            LL ans1 = 0, ans2 = 1e9;            int s = i;            for(int j = 0; j <= 35; j++) {                if(k & f[j]) {                    ans1 += sum[s][j];                    ans2 = min(ans2, Min[s][j]);                    s = g[s][j];                }            }            printf("%lld %lld\n", ans1, ans2);        }    }    return 0;}

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