CF750C

New Year and Rating

题目链接

题目

样例

输入

3
-7 1
5 2
8 2

输出

1907

思路

二分

二分rating，从1到n遍历，若碰到不满足条件的：1却max小于1900，2却min大于1899，则直接返回，修改mid的区间重新取
若满足条件，则直接加减所给值

模拟

和上面一样，若不满足直接输出，满足则一直加

代码

点击查看代码

#include <iostream>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <string>
#include <stack>
#include <queue>
#include <unordered_map>
#include <map>
#include <vector>
#include <cstring>
#include <bitset>
#include <set>

using namespace std;
using ll = long long;
ll ma = 2e8 + 10;
int mi;

int gcd(int a, int b) {return b == 0 ? a : gcd(b, a % b);}
int lcm(int a, int b) {return a / gcd(a, b) * b;}

void solve()
{
    int n; cin >> n;
    while(n--) {
    	int a, b; cin >> a >> b;
    	if((b == 1 && ma < 1900) || (b == 2 && mi > 1899)) {
    		cout << "Impossible";
    		return;
    	}
    	if(b == 1 && mi < 1900) mi = 1900;
    	if(b == 2 && ma > 1899) ma = 1899;
    	if(mi != 0) mi += a;
    	if(ma != 2e8 + 10) ma += a;
    }
    if(ma == 2e8 + 10) cout << "Infinity";
    else cout << ma;
}

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    //int t; cin >> t;
    int t = 1;
    while (t--) solve();
}