Problem L — limit 1 second
Liars
There are n people in a circle, numbered from 1 to n, each of whom always tells the truth or always lies. Each person i makes a claim of the form: “the number of truth-tellers in this circle is between ai and bi , inclusive.” Compute the maximum number of people who could be telling the truth. Input The first line contains a single integer n (1 ≤ n ≤ 103 ). Each of the next n lines contains two space-separated integers ai and bi (0 ≤ ai ≤ bi ≤ n). Output Print, on a single line, the maximum number of people who could be telling the truth. If the given set of statements is inconsistent, print -1 instead.
Sample Input and Output 3 1 1 2 3 2 2 2 8 0 1 1 7 4 8 3 7 1 2 4 5 3 7 1 8 -1
用差分数组对所有说真话人数的可能性进行计数，之后暴力求最大值
//#include <bits/stdc++.h> #include <iostream> #include <cstdio> #include <cmath> #include<cstring> #include <algorithm> #include <queue> #include<map> #include<set> #include<vector> using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int mod = 1e9; const int mx = 1e7; //check the limits, dummy typedef pair<int, int> pa; const double PI = acos(-1); ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; } #define swa(a,b) a^=b^=a^=b #define re(i,a,b) for(int i=(a),_=(b);i<_;i++) #define rb(i,a,b) for(int i=(a),_=(b);i>=_;i--) #define clr(a) memset(a, inf, sizeof(a)) #define lowbit(x) ((x)&(x-1)) #define mkp make_pai //void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); } int n, m,t; int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); cin >> n; vector<int>dif(n + 1, 0); for (int i = 0, l, r; i < n; i++) { cin >> l >> r; dif[l]++; if (r != n)dif[r + 1]--; } vector<int>cnt(n + 1, 0); cnt[0] = dif[0]; int ans = -1; re(i, 1, n + 1) { cnt[i] = cnt[i - 1] + dif[i]; if (cnt[i] == i)ans = i; } cout << ans << endl; return 0; }