【codeforces 766C】Mahmoud and a Message
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Mahmoud wrote a message s of length n. He wants to send it as a birthday present to his friend Moaz who likes strings. He wrote it on a magical paper but he was surprised because some characters disappeared while writing the string. That’s because this magical paper doesn’t allow character number i in the English alphabet to be written on it in a string of length more than ai. For example, if a1 = 2 he can’t write character ‘a’ on this paper in a string of length 3 or more. String “aa” is allowed while string “aaa” is not.
Mahmoud decided to split the message into some non-empty substrings so that he can write every substring on an independent magical paper and fulfill the condition. The sum of their lengths should be n and they shouldn’t overlap. For example, if a1 = 2 and he wants to send string “aaa”, he can split it into “a” and “aa” and use 2 magical papers, or into “a”, “a” and “a” and use 3 magical papers. He can’t split it into “aa” and “aa” because the sum of their lengths is greater than n. He can split the message into single string if it fulfills the conditions.
A substring of string s is a string that consists of some consecutive characters from string s, strings “ab”, “abc” and “b” are substrings of string “abc”, while strings “acb” and “ac” are not. Any string is a substring of itself.
While Mahmoud was thinking of how to split the message, Ehab told him that there are many ways to split it. After that Mahmoud asked you three questions:
How many ways are there to split the string into substrings such that every substring fulfills the condition of the magical paper, the sum of their lengths is n and they don’t overlap? Compute the answer modulo 109 + 7.
What is the maximum length of a substring that can appear in some valid splitting?
What is the minimum number of substrings the message can be spit in?
Two ways are considered different, if the sets of split positions differ. For example, splitting “aa|a” and “a|aa” are considered different splittings of message “aaa”.
Input
The first line contains an integer n (1 ≤ n ≤ 103) denoting the length of the message.
The second line contains the message s of length n that consists of lowercase English letters.
The third line contains 26 integers a1, a2, …, a26 (1 ≤ ax ≤ 103) — the maximum lengths of substring each letter can appear in.
Output
Print three lines.
In the first line print the number of ways to split the message into substrings and fulfill the conditions mentioned in the problem modulo 109 + 7.
In the second line print the length of the longest substring over all the ways.
In the third line print the minimum number of substrings over all the ways.
Examples
input
3
aab
2 3 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
output
3
2
2
input
10
abcdeabcde
5 5 5 5 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
output
401
4
3
Note
In the first example the three ways to split the message are:
a|a|b
aa|b
a|ab
The longest substrings are “aa” and “ab” of length 2.
The minimum number of substrings is 2 in “a|ab” or “aa|b”.
Notice that “aab” is not a possible splitting because the letter ‘a’ appears in a substring of length 3, while a1 = 2.
【题目链接】:http://codeforces.com/contest/766/problem/C
【题意】
给你一个长度为n的字符串;
这个字符串只包含小写字母;
然后让你把这个字符串进行分割;形成若干个小的字符串;
但是不是任意分割的;
每个小写字母都有一个数字ma[i];表示这个字母能够存在于长度不超过ma[i]的字符串内;
在这个条件下分割;
【题解】
设f[i]表示从i开始进行分割的方案数;
f[i] += ∑f[j];这里min(ma[s[i]..s[j]])>=j-i+1,且j>=i;
ma[x]是x这个字母能够待在的最长的字符串的长度;
然后每次都用j-i+1尝试更新“段”的最大值;
用一个num[i]表示以i作为分割的起点需要分成几段;
num[i]=min(num[i],num[j]+1);
边界:
f[n+1]=1,num[n+1]=0;
逆序更新;
最后输出f[1]就好;
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1100;
const int MOD = 1e9+7;
int n,ma[MAXN],a[MAXN],mal = 0,mi,num[MAXN];
LL ans = 0,f[MAXN];
char s[MAXN];
void solve(int x)
{
int change = 1e8;
rep1(i,x,n)
{
change = min(ma[a[i]],change);
int ll = i-x+1;
if (ll>change)
break;
if (f[i+1]!=-1)
{
if (f[x]==-1) f[x] = 0;
f[x] = (f[x]+f[i+1])%MOD;
num[x] = min(num[i+1]+1,num[x]);
mal = max(mal,ll);
}
}
}
int main()
{
//freopen("F:\\rush.txt","r",stdin);
memset(f,255,sizeof f);
memset(num,0x3f3f3f3f,sizeof num);
rei(n);
scanf("%s",s+1);
rep1(i,1,n)
a[i] = s[i]-'a'+1;
rep1(i,1,26)
rei(ma[i]);
f[n+1] = 1,num[n+1] = 0;
rep2(i,n,1)
solve(i);
mi = n;
cout << f[1] << endl;
printf("%d\n",mal);
printf("%d\n",num[1]);
return 0;
}
【codeforces 766C】Mahmoud and a Message的更多相关文章
- 【codeforces 766D】Mahmoud and a Dictionary
time limit per test4 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 766A】Mahmoud and Longest Uncommon Subsequence
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 766B】Mahmoud and a Triangle
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- 【codeforces 766E】Mahmoud and a xor trip
[题目链接]:http://codeforces.com/contest/766/problem/E [题意] 定义树上任意两点之间的距离为这条简单路径上经过的点; 那些点上的权值的所有异或; 求任意 ...
- Codeforces 766C:Mahmoud and a Message(DP)
题目链接:http://codeforces.com/problemset/problem/766/C 题意 有一个长度为n的字符串,第二行有26个数字,位置1~26对应为a~z的字母,数值表示该字母 ...
- 【codeforces 415D】Mashmokh and ACM(普通dp)
[codeforces 415D]Mashmokh and ACM 题意:美丽数列定义:对于数列中的每一个i都满足:arr[i+1]%arr[i]==0 输入n,k(1<=n,k<=200 ...
- 【codeforces 707E】Garlands
[题目链接]:http://codeforces.com/contest/707/problem/E [题意] 给你一个n*m的方阵; 里面有k个联通块; 这k个联通块,每个连通块里面都是灯; 给你q ...
- 【codeforces 707C】Pythagorean Triples
[题目链接]:http://codeforces.com/contest/707/problem/C [题意] 给你一个数字n; 问你这个数字是不是某个三角形的一条边; 如果是让你输出另外两条边的大小 ...
- 【codeforces 709D】Recover the String
[题目链接]:http://codeforces.com/problemset/problem/709/D [题意] 给你一个序列; 给出01子列和10子列和00子列以及11子列的个数; 然后让你输出 ...
随机推荐
- vue_qqmapdemo1
腾讯地图vue组件,实现异步加载腾讯地图,坐标拾取器,支持按城市名称搜索. 搜索框样式依赖elementUI,不需要可删除顶部,地图部分无依赖项 //qqmap.vue <template> ...
- 让超出div内容的显示滚动条:overflow:auto,以及overflow其它属性
css的属性,以前没用过遇到了,记录一下: 虽然layui本来自带这个处理,但是为了灵活,抛弃layui原有的加载,只是用layui的样样式,就要使用到这个css属性 总结overflow属性: /* ...
- Ui自动化测试框架
为了提高我们的UI测试效率,我们引用Ui自动化测试框架,这里简单先描述一下,后续会详细补充: 了解一个测试框架,我们就需要了解一下源码,能看懂源码即可: 1.稳定先封装wait EC,电脑性能配置较好 ...
- laravel 操作日志;
在网上寻找了许多方法,觉得有的地方看不懂, 决定自己写一些关于laravel中调用本身中的操作日志: Laravel 日志工具在强大的 Monolog 函数库上提供一层简单的功能.Laravel 默 ...
- 【JZOJ4868】【NOIP2016提高A组集训第9场11.7】Simple
题目描述 数据范围 解法 在暴力枚举的基础上,当n的系数在[0,m/gcd(n,m))时,得到的c是不重复不遗漏的. 设n的系数为x,m的系数为y. 不重复不遗漏性 设x=m/gcd(n,m)+i,那 ...
- AtCoder Beginner Contest 084 C - Special Trains
Special Trains Problem Statement A railroad running from west to east in Atcoder Kingdom is now comp ...
- Android 在图片的指定位置添加标记
这些天,项目里加了一个功能效果,场景是: 假如有一个家居图片,图片里,有各样的家居用品: 桌子,毛巾,花瓶等等,需要在指定的商品处添加标记,方便用户直接看到商品,点击该标记,可以进入到商品详情页 .实 ...
- JDK8中`Optional.orElse()` 和`Optional.orElseGet()`之间的区别
看例子就明白了 ``` static String B() { System.out.println("B()..."); return "B"; } publ ...
- MySQL列出当前月的每一天
因为工作的原因,要用MySQL列出当前月份每一天的日期,自己查了下网上资料都是列出最近一个月的日期的解决方案,自己根据查到的的方案,修改成了下面两个方案,在此记录下: 方案一: SELECT date ...
- oracle函数 SUBSTRB(c1,n1[,n2])
[功能]取子字符串 [说明]多字节符(汉字.全角符等),按2个字符计算 [参数]在字符表达式c1里,从n1开始取n2个字符;若不指定n2,则从第y个字符直到结束的字串. [返回]字符型,如果从多字符右 ...