CodeForces - 750C（约束上下界）

---

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.

Example
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.

---

给出打每场比赛前的等级和这场比赛的得分情况，要求求出最终总分数的最大值，因为等级一有下界1900，等级二有上界1899，所以初始化初始得分的上下界后，每一次比赛前的等级就可以帮助我们根据之前的得分情况约束上界或者下界，如果位于等级1就约束下界，位于等级二就约束上界，最终如果上界没被约束过答案就是inf，如果下界比上界还大就是impossible，不然就输出上界+每场得分。

#include<iostream>#include<stdio.h>#include<algorithm>#include<string.h>#include<string>#include<map>#include<set>#include<vector>using namespace std;int dd[200005];int l[200005];int n;int main(){    scanf("%d",&n);    for(int i=0;i<n;i++){        scanf("%d%d",&dd[i],&l[i]);    }    int up,down;    if(l[0]==2){down=-30000000;up=1899;}    else{down=1900;up=30000000;}    int cnt=0;    for(int i=1;i<n;i++){        cnt+=dd[i-1];        if(l[i]==1){            if(down+cnt<1900){down=1900-cnt;}        }        else{            if(up+cnt>1899){up=1899-cnt;}        }    }    int ans;    if(up==30000000){printf("Infinity\n");}    else if(down>up){printf("Impossible\n");}    else{        ans=up;        for(int i=0;i<n;i++){ans+=dd[i];}        printf("%d\n",ans);    }return 0;}