POJ 3484 - Showstopper(二分搜索)

来源：互联网发布：java调用接口传递参数编辑：程序博客网时间：2024/05/22 08:29

题目:

http://poj.org/problem?id=3484

思路:

又是二分咯, 这道题的输入比较神奇,以换行符来判断一个样例的结束.用到sscanf.

AC.

#include <iostream>#include <cstdio>#include <cstring>using namespace std;typedef long long ll;const int maxn = 1000005;char s[100];int n;ll x[maxn], y[maxn], z[maxn];bool can(ll m){    ll s = 0;    for(int i = 0; i < n; ++i) {        if(m < x[i]) continue;        else if(m > y[i]) s += ( (y[i]-x[i])/z[i]+1 );        else s += ( (m - x[i])/z[i]+1 );    }    if(s % 2) {        return true;    }    return false;}void solve(){    ll l = 0, r = 1e15;    while(r - l > 1) {        ll mid = (r+l)/2;        if(can(mid)) r = mid;        else l = mid;    }    if(l == 1e15-1) printf("no corruption\n");    else {        ll sum = 0;        for(int i = 0; i < n; ++i) {            if(r >= x[i] && r <= y[i] && (r - x[i]) % z[i] == 0)            sum++;        }        printf("%I64d %I64d\n", r, sum);    }}int main(){//freopen("in", "r", stdin);    n = 0;    while(gets(s) != NULL) {        if(strlen(s) != 0) {            sscanf(s, "%lld %lld %lld", &x[n], &y[n], &z[n]);            n++;        }        else if(n) {            solve();            n = 0;        }    }    if(n) {        solve();    }    return 0;}

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