Codeforces 876 B Divisiblity of Differences 基础数学
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题意
给出n,k,m(100000),n个数(1e9),从中选出k个数使得任意两个数的差为m.
解法
全取模,如果模的一个结果的个数>=k,就用它.全没超过,则不可能.
其他
重写盲交错误:
1.未定义ans.—–使用先定义.
2.两次使用i——-索引变量具体名字,改习惯
3.i与have[i]搞混–索引变量具体名字,改习惯.
代码
/* LittleFall : Hello! */#include <bits/stdc++.h>using namespace std;#define M 100016int save[M],have[M];int main(void){ int n,k,m; scanf("%d%d%d",&n,&k,&m); for(int i=0;i<n;i++) { scanf("%d",&save[i]); have[save[i]%m]++; } int ans=0; for(int i=0;i<m;i++) { if(have[i]>=k) { printf("Yes\n"); for(int j=0;j<n;j++) { if(save[j]%m==i) { printf("%d ",save[j] ); ans++; if(ans==k) return 0; } } } } printf("No\n"); return 0;}
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