标签：nec 奇数 图片 分享 get 观察 freopen lib cti
题意：
有N个社团，每个社团三个属性A,B,C，表示会向编号A+k*C的同学发传单(k=0,1,2... && A+k*C <= B)。题目保证最多有一个人收到的传单数是奇数。求如果有人收到传单是奇数，输出编号和他收到的传单数。
思路：
观察最后情况，可以发现，要么每个人都是偶数。要么有一个是奇数。观察其前缀和，发现奇数那个人之前的前缀和都是偶数，之后都是奇数。所以，二分之。
代码：
（上交大牛代码……）
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
int n, a[100001][3];
long long sum;
bool check(int now){
long long sum = 0;
for (int i = 1; i <= n; i++)
if (now >= a[i][0])
{
int Right = min(a[i][1], now);
if (a[i][2]) sum += (long long)(Right - a[i][0]) / a[i][2] + 1;
else sum++;
}
return(sum & 1);
}
int read(){
char ch;
for (ch = getchar(); ch < ‘0‘ || ch > ‘9‘; ch = getchar());
int cnt = 0;
for (; ch >= ‘0‘ && ch <= ‘9‘; ch = getchar()) cnt = cnt * 10 + ch - ‘0‘;
return(cnt);
}
int main(){
//freopen("j.in", "r", stdin);
//freopen("j.out", "w", stdout);
for (;;)
{
if (scanf("%d", &n) != 1) return 0;
sum = 0;
int Max = 0;
for (int i = 1; i <= n; i++)
{
a[i][0] = read(); a[i][1] = read(); a[i][2] = read();
if (a[i][2]) sum += (long long)(a[i][1] - a[i][0]) / a[i][2] + 1;
else sum++;
Max = max(Max, a[i][1]);
}
if (!(sum & 1))
{
printf("DC Qiang is unhappy.\n");
continue;
}
long long Left = 0, Right = Max, Mid = (Left + Right) >> 1;
while (Left + 1 < Right)
{
if (check(Mid)) Right = Mid;
else Left = Mid;
Mid = (Left + Right) >> 1;
}
printf("%d", Right);
int cnt = 0;
for (int i = 1; i <= n; i++)
if (Right >= a[i][0] && Right <= a[i][1] && !((Right - a[i][0]) % a[i][2])) ++cnt;
printf(" %d\n", cnt);
}
}
标签：nec 奇数 图片 分享 get 观察 freopen lib cti
原文地址：http://www.cnblogs.com/unknownname/p/7903575.html
