$CF1153A\ Serval\ and\ Bus$

$CF1153A\ Serval\ and\ Bus$

看大佬的代码都好复杂(不愧是大佬\(orz\)

蒟蒻提供一种思路

因为求的是最近的车对吧\(qwq\)

所以我们可以用一个\(while\)循环所以没必要去用什么 \(for...\)

至于这是\(div2\)的第一题还是比较水的

#include <bits/stdc++.h>
#define rep(i,j,n) for(register int i=j;i<=n;i++)
#define Rep(i,j,n) for(register int i=j;i>=n;i--)
#define low(x) x&(-x)
using namespace std ;
typedef long long LL ;
const int inf = INT_MAX >> 1 ;
inline LL In() { LL res(0) , f(1) ; register char c ;
#define gc c = getchar()
    while(isspace(gc)) ; c == '-' ? f = - 1 , gc : 0 ;
    while(res = (res << 1) + (res << 3) + (c & 15) , isdigit(gc)) ;
    return res * f ;
#undef gc
}

int n , t ;
const int N = 100 + 5 ;
struct node {
    int s , d ;
}a[N] ;
inline void Ot() {
    n = In() , t = In() ;
    rep(i,1,n) a[i].s = In() , a[i].d = In() ;
    int tmp = inf ;
    int ans = 0 ;
    rep(i,1,n) {
        int s = a[i].s ;
        while(s < t) s += a[i].d ;
        if(tmp > (s-t)) tmp = (s-t) , ans = i ;
    }
    cout << ans << endl ;
}
signed main() {
//  freopen("test.in","r",stdin) ;
    return Ot() , 0 ;
}
posted @ 2019-04-19 14:45 breezeღ 阅读( ...) 评论( ...) 编辑 收藏

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转载自blog.csdn.net/qq_42628055/article/details/89410748