前言

高三dalao试图混入其中

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成绩

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$1$	$2017myself$	$205$	$25$	$80$	$100$
$2$	$2013lyy$	$200$	$100$	$70$	$30$
$3$	$2015hjw$	$120$	$30$	$90$	$0$
$4$	$2017zyc$	$110$	$30$	$30$	$50$
$4$	$2013lyk$	$110$	$100$	$10$	$0$
$6$	$2015trx$	$105$	$5$	$100$	$0$
$7$	$2017xxy$	$100$	$100$	$0$	$0$
$7$	$2017xjq$	$100$	$0$	$100$	$0$
$9$	$2015gjh$	$70$	$0$	$70$	$0$
$10$	$2013yhl$	$50$	$0$	$0$	$50$

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正题

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$T1:nssl1231-Gift$【$01$背包$,dp$】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/83445478

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$T2:nssl1232-$函数【数论,欧拉函数$,$莫比乌斯反演】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/83445596

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$T3:nssl1230-$序列【位运算】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/83445650

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$someofcode$

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T1 25分code

#include<cstdio>
#include<algorithm>
using namespace std;
int n,m,c[100],s;
int main()
{scanf("%d%d",&n,&m);for(int i=1;i<=n;i++)scanf("%d",&c[i]);for(int i=0;i<(1<<n);i++){int mo=0,mins=2147483647;for(int j=0;j<n;j++){mo+=(i>>j)%2*c[j+1];if(!((i>>j)%2))mins=min(mins,c[j+1]);}if(mo<=m&&mo+mins>m) s++;}printf("%d",s);
}

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尾声

用更高深的算法水了个80分？