time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

Every Codeforces user has rating, described with one integer, possibly negative or zero. Users are divided into two divisions. The first division is for users with rating 1900 or higher. Those with rating 1899 or lower belong to the second division. In every contest, according to one’s performance, his or her rating changes by some value, possibly negative or zero.

Limak competed in n contests in the year 2016. He remembers that in the i-th contest he competed in the division di (i.e. he belonged to this division just before the start of this contest) and his rating changed by ci just after the contest. Note that negative ci denotes the loss of rating.

What is the maximum possible rating Limak can have right now, after all n contests? If his rating may be arbitrarily big, print “Infinity”. If there is no scenario matching the given information, print “Impossible”.

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000).

The i-th of next n lines contains two integers ci and di ( - 100 ≤ ci ≤ 100, 1 ≤ di ≤ 2), describing Limak’s rating change after the i-th contest and his division during the i-th contest contest.

Output

If Limak’s current rating can be arbitrarily big, print “Infinity” (without quotes). If the situation is impossible, print “Impossible” (without quotes). Otherwise print one integer, denoting the maximum possible value of Limak’s current rating, i.e. rating after the n contests.

Examples

input

3

-7 1

5 2

8 2

output

1907

input

2

57 1

22 2

output

Impossible

input

1

-5 1

output

Infinity

input

4

27 2

13 1

-50 1

8 2

output

1897

Note

In the first sample, the following scenario matches all information Limak remembers and has maximum possible final rating:

Limak has rating 1901 and belongs to the division 1 in the first contest. His rating decreases by 7.

With rating 1894 Limak is in the division 2. His rating increases by 5.

Limak has rating 1899 and is still in the division 2. In the last contest of the year he gets  + 8 and ends the year with rating 1907.

In the second sample, it’s impossible that Limak is in the division 1, his rating increases by 57 and after that Limak is in the division 2 in the second contest.

【题目链接】:http://codeforces.com/contest/750/problem/C

【题解】



我用的应该是最蠢的办法了;

敲了180行.

敲到一半我都快要哭出来了。

但是痛苦是有回报的:)

这次代码的注解写的很详细.可以到程序里面具体看;

/*
如果一直都没有出现div2
那就说明全都在div1;
则不管你怎么变都无所谓,都是无限大
如果全都是div2他有一个上限1900
(有可能从div2变成了div1,所以要特判一下)
x=0
然后对x进行ci的变换
记录x的最大值
如果max(x)>=0
则rating为
1899-(max(x)-x)
如果max(x)<0
则rating为
1899+x 有div1和div2夹杂
找最后一个div1和div2的分割点
i c[i] div2
i+1 c[i+1] div1
那就表示第i场比赛的时候
rating<1900
第i+1场比赛的时候rating>1900
ci变化范围是1..100
所以第i场的时候分数范围是
1800..1899
从大到小枚举
ci不在这个范围输出无解 i c[i] div1
i+1 c[i+1] div2
则第i场比赛的时候
rating>=1900
第i+1场比赛的时候rating<1900
第i场比赛的时候rating的范围是1900..1999
ci的话范围是-1..-100
如果ci不在这个范围则也输出无解
时间复杂度就是O(100*N)吧。不会超的.
*/

【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 MAXN = 2e5+100;
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); struct abc
{
int c,d;
}; int n,nd[3];
abc a[MAXN]; int main()
{
//freopen("F:\\rush.txt","r",stdin);
rei(n);
rep1(i,1,n)
{
rei(a[i].c);rei(a[i].d);
nd[a[i].d]++;
}
if (nd[1]>0 && nd[2]==0)//都是div1
{
puts("Infinity");
return 0;
}
if (nd[1]==0 && nd[2]>0)//都是div2
{
rep2(rating,1999,1900)//最后变成div1的情况
{
int tempr = rating;
bool fi = true;
rep2(j,n,1)
{
tempr+=-a[j].c;
if (tempr>=1900)
{
fi = false;
break;
}
}
if (!fi)
continue;
if (fi)
{
printf("%d\n",rating);
return 0;
}
}
int x = 0,ma = -2100000000,ans;//其他则都在div2
rep1(i,1,n)
{
x+=a[i].c;
ma = max(ma,x);
}
if (ma>=0)//最大值大于0那个时候到达最大1899
ans = 1899-(ma-x);
else//最大值小于0,则总是小于1899的
ans = 1899+x;
cout << ans << endl;
return 0;
}
rep2(i,n,2)
if (a[i].d!=a[i-1].d)
{
if (a[i-1].d==2 && a[i].d==1)//之前是div2涨分到div1
{
rep2(rating,1899,1800)//从大到小枚举打i-1这场比赛之前的分数
{
bool fi = true;
int tempr = rating;
rep2(j,i-2,1)
{
tempr+=(-a[j].c);//看看减回去之后能不能打那场比赛
if (tempr>=1900 && a[j].d==2)
{
fi = false;
break;
}
if (tempr<1900 && a[j].d==1)
{
fi = false;
break;
}
}
if (!fi) continue;
tempr = rating;
rep1(j,i-1,n)//看看打完这场比赛之后是不是真的能打下一场比赛
{
tempr+=a[j].c;
if (tempr>=1900 && a[j+1].d==2)//判断依据是下一场比赛
{
fi = false;
break;
}
if (tempr<1900 && a[j+1].d==1)
{
fi = false;
break;
}
}
if (!fi) continue;
if (fi)
{
printf("%d\n",tempr);
return 0;
}
}
puts("Impossible");
return 0;
}
else
if (a[i-1].d==1 && a[i].d==2)
{
rep2(rating,1999,1900)//打第i-1场比赛之前分数为1900.1999这个区间
{
bool fi = true;
int tempr = rating;
rep2(j,i-2,1)//看看倒回去是不是每场比赛都真的能打
{
tempr+=(-a[j].c);
if (tempr>=1900 && a[j].d==2)
{
fi = false;
break;
}
if (tempr<1900 && a[j].d==1)
{
fi = false;
break;
}
}
if (!fi) continue;
tempr = rating;
rep1(j,i-1,n)//继续往前打
{
tempr+=a[j].c;
if (tempr>=1900 && a[j+1].d==2)//看看下一场比赛能不能打
{
fi = false;
break;
}
if (tempr<1900 && a[j+1].d==1)
{
fi = false;
break;
}
}
if (!fi) continue;
if (fi)
{
printf("%d\n",tempr);
return 0;
}
}
puts("Impossible");
return 0;
}
break;
}
return 0;
}

【codeforces 750C】New Year and Rating的更多相关文章

  1. 【codeforces 750C】New Year and Rating(做法2)

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  2. 【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 ...

  3. 【codeforces 807A】Is it rated?

    [题目链接]:http://codeforces.com/contest/807/problem/A [题意] 给你n个人在一场CF前后的rating值; 问你这场比赛是不是计分的 [题解] 如果有一 ...

  4. 【74.89%】【codeforces 551A】GukiZ and Contest

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  5. 【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 ...

  6. 【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 ...

  7. 【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 ...

  8. 【codeforces 709B】Checkpoints

    [题目链接]:http://codeforces.com/contest/709/problem/B [题意] 让你从起点开始走过n-1个点(至少n-1个) 问你最少走多远; [题解] 肯定不多走啊; ...

  9. 【codeforces 709C】Letters Cyclic Shift

    [题目链接]:http://codeforces.com/contest/709/problem/C [题意] 让你改变一个字符串的子集(连续的一段); ->这一段的每个字符的字母都变成之前的一 ...

随机推荐

  1. [React Native]升级React Native版本

    React Native正式版本还没发布,但是小版本基本上每个月都更新1-2次.9月11号又更新了0.33版本,其中有两个增强功能正好是项目中用到的. 添加Android6.0权限验证API Add ...

  2. mysql字段中提取汉字,去除数字以及字母

    如果只是删除尾部的中文,保留数据,可以用以下的简单方式 MySQL as num; +------+ | num | +------+ | +------+ DELIMITER $$ DROP FUN ...

  3. CTR+A组合键 以及终止按键事件传递

    Key UP 或Down 事件中 实现CTR+A全选 if ( Control.ModifierKeys==Keys.Control && e.KeyCode == Keys.A)   ...

  4. Redis源码解析:07压缩列表

    压缩列表(ziplist)是列表键和哈希键的底层实现之一.当列表键只包含少量列表项,并且每个列表项要么是小整数值,要么是长度较短的字符串时:或者当哈希键只包含少量键值对,并且每个键值对的键和值要么是小 ...

  5. 在SpringBoot中使用JWT

    JWT简介 简介 JSON Web token简称JWT, 是用于对应用程序上的用户进行身份验证的标记.也就是说, 使用 JWTS 的应用程序不再需要保存有关其用户的 cookie 或其他sessio ...

  6. oracle 减少对表的查询

    在含有子查询的SQL语句中,要特别注意减少对表的查询. 例如: 低效 SELECT TAB_NAME FROM TABLES WHERE TAB_NAME = ( SELECT TAB_NAME FR ...

  7. 2013-2-1 pdf中无法用金山词霸取词问题

    打开pdf的编辑——〉首选项——〉一般——〉选项——〉开始——〉只有经过认证的插件,把‘checkbox’里的勾去掉,重启. ★在acrobat reader启动画面里如果没有加载xdict32(工具 ...

  8. ubuntu14.04 编译hadoop-2.6.0-cdh5.4.4

    1 protocol buffer sudo apt-get install libprotobuf-dev asn@hadoop1:~/Desktop$ protoc --version libpr ...

  9. Python第三方包的egg info 是什么东西

    xxx.egg-info 一般与 xxx文件夹同时存在,一起来表示完整模块.

  10. SuperSocket命令和命令加载器

    关键字: 命令, 命令加载器, 多命令程序集 命令 (Command) SuperSocket 中的命令设计出来是为了处理来自客户端的请求的, 它在业务逻辑处理之中起到了很重要的作用. 命令类必须实现 ...