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

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】

设一开始的时候分数为x
则如果第一个d = 1;
则上限为INF
如果第一个d=2
则上限为1899
如果下一场比赛为div1
则x+c1+c2+..+ci>=1900
如果下一场比赛为div2
则x+c1+c2+..+ci<=1899
即
x>=1900-c1-c2..ci div1
x<=1899-c1-c2..ci; div2
在做的过程中搞一个前缀和
delta+=ci;
(delta=c1+c2..ci);
维护一下x的最小值和最大值;
然后用x的最大值加上delta;
就能直接获取最后的分数了;
最小值如果大于最大值，则不合法;
最大值如果为Inf就表示全是div1;
一开始的时候可以弄一个技巧
就是delta的值后加上ci;
这样delta就保留了上一次的变化;
这样就不用错开一位往后判断了;

【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 INF = 1e9;
const int MAXN = 2e5+10;
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);

int n,delta,ma,mi;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ma = INF,mi = -INF,delta = 0;
    rei(n);
    rep1(i,1,n)
    {
        int x,y;
        rei(x);rei(y);
        if (y==1)
            mi = max(mi,1900-delta);
        else
            ma = min(ma,1899-delta);
        delta+=x;
    }
    if (mi>ma)
    {
        puts("Impossible");
        return 0;
    }
    if (ma==INF)
    {
        puts("Infinity");
        return 0;
    }
    printf("%d\n",ma+delta);
    return 0;
}

转载于:https://www.cnblogs.com/AWCXV/p/7626743.html