2017 Multi-University Training Contest 7 hdu 6124

HDU - 6124

题意:给定正整数aa,求对于所有正整数bb,amod bamodb有多少种可能的结果。1leq aleq10^91a109​​。

思路:显然小于<a/2的每个非负整数c都是可能成为余数的,取b=a-c即可,另外a本身也能成为余数,而其他数显然都不可能。

AC代码:

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("
")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const long long INF = 1e18+1LL;
const int inf = 1e9+1e8;
const int N=1e7+100;
const ll mod=1e9+7;
const double PI=acos(-1.0);

int main(){
    ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
    int T; cin>>T;
    while(T--){
        int a,d=0; cin>>a;
        cout<<(a+1>>1)+1<<endl;
    }
    return 0;
}
原文地址:https://www.cnblogs.com/max88888888/p/7367934.html