Educational Codeforces Round 8 ABCDE

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套题链接：http://codeforces.com/contest/628

难度类型：前几题比较容易，D题后难度陡增。

A

题解

类型：模拟

根据题意一轮轮做就是了，只是题目有点长，要仔细读。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <stack>#include <queue>#include <string>#include <vector>#include <set>#include <map>#define fi first#define se secondusing namespace std;typedef long long LL;typedef pair<int,int> PII;// headint get(int x) {    for (int i = 31; ~i; i--) {        if ((1<<i) & x) return (1<<i);    }}int main() {    int n, b, p;    scanf("%d%d%d", &n, &b, &p);    int ansb = 0, ansp = p * n;    while (n > 1) {        int x = get(n);        ansb += b*x + x/2;        n = x/2 + n - x;    }    printf("%d %d\n", ansb, ansp);    return 0;}

B

题解

类型：dp，递推

需要注意是要连续的，然后就是进行一个线性的递推就行了。

dp数组的意义是以某位置为结尾，模4为几的个数。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <stack>#include <queue>#include <string>#include <vector>#include <set>#include <map>#define fi first#define se secondusing namespace std;typedef long long LL;typedef pair<int,int> PII;// headconst int N = 3e5+5;const int mod = 4;char s[N];LL dp[2][mod];int main() {    gets(s);    int n = strlen(s);    int cur = 1, pre = 0;    LL ans = 0;    for (int i = 0; i < n; i++) {        int v = s[i] - '0';        for (int j = 0; j < mod; j++) {            dp[cur][j] = 0;        }        for (int j = 0; j < mod; j++) {            int nxt = (j * 10 + v) % mod;            dp[cur][nxt] += dp[pre][j];        }        dp[cur][v%mod]++;        ans += dp[cur][0];        swap(cur, pre);    }    printf("%I64d\n", ans);    return 0;}

C

题解

类型：贪心，模拟

要求变化距离的总和为某数，那尽量多去加，剩下的不变就行了。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#include <cstring>#include <stack>#include <queue>#include <string>#include <vector>#include <set>#include <map>#define fi first#define se secondusing namespace std;typedef long long LL;typedef pair<int,int> PII;// headconst int N = 1e5+5;char s[N];int main() {    int n, k;    scanf("%d%d%s", &n, &k, s);    for (int i = 0; i < n && k; i++) {        bool up = abs(s[i] - 'a') < abs(s[i] - 'z');        int mx = max(abs(s[i] - 'a'), abs(s[i] - 'z'));        mx = min(mx, k);        s[i] += up ? mx : -mx;        k -= mx;    }    puts(k ? "-1" : s);    return 0;}

D

题解

类型：数位dp

传送门：http://blog.csdn.net/xc19952007/article/details/50720986

E

题解

类型：数据结构，技巧

传送门：http://blog.csdn.net/xc19952007/article/details/50721057

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