前言

成功翻车

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成绩

$Rank$是有算别人的

$Rank$	$Person$	$Score$	$A$	$B$	$C$
$14$	$2017hzb$	$80$	$30$	$0$	$50$
$14$	$2017wyc$	$80$	$0$	$30$	$50$
$14$	$2017xxy$	$80$	$30$	$0$	$50$
$44$	$2017lw$	$30$	$30$	$0$	$0$
$44$	$2017lrz$	$30$	$30$	$0$	$0$
$44$	$2017zyc$	$30$	$0$	$30$	$0$
$44$	$2017hjq$	$30$	$0$	$0$	$30$
$61$	$2017xjq$	$0$	$0$	$0$	$0$

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正题

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$T1:jzoj4208-$线段树什么的最讨厌了【$dfs$】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86577063

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$T2:jzoj4209-$已经没有什么好害怕的了【差分】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86577360

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$T3:jozj4010-$我才不是萝莉控呢【哈夫曼树】

博客链接:
https://blog.csdn.net/Mr_wuyongcong/article/details/86577658

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$someofcode$

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T2暴力

#include<cstdio>
#include<cstring>
#define ll long long
using namespace std;
const ll XJQ=1e9+7;
char s[1001];
ll n,ans[1001],sum,t;
int main()
{freopen("data.in","r",stdin);freopen("data.ans","w",stdout);scanf("%lld",&t);while(t--){scanf("%s",s+1);n=strlen(s+1);memset(ans,0,sizeof(ans));sum=0;for(ll i=1;i<n;i++){for(ll j=i+1;j<=n;j++){ll left=0;bool flag=1;for(ll k=i;k<=j;k++){if(s[k]=='(') left++;else left--;if(left<0){flag=0;break;}}if(flag&&!left)for(ll k=i;k<=j;k++)ans[k]++;}}for(ll i=1;i<=n;i++)sum+=ans[i]*i%XJQ;printf("%lld\n",sum);}
}

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T2随机数据

#include<cstdio>
#include<cstdlib>
#include<ctime>
using namespace std;
int t,n;
int main()
{freopen("data.in","w",stdout);srand(time(0));t=10;n=100;printf("10\n");for(int i=1;i<=t;i++){//for(int j=1;j<=n/2;j++)// printf("(");for(int j=1;j<=n;j++)// printf(")");if(rand()&1) printf(")");else printf("(");printf("\n");}
}

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T2 30code

#include<cstdio>
#include<cstring>
#include<stack>
#define ll long long
using namespace std;
const ll N=1e6+10,XJQ=1e9+7;
stack<ll> left,mark,k;
ll t,n,ans,sum[N],ks,marks;
char s[N];
int main()
{freopen("data.in","r",stdin);freopen("data.out","w",stdout);scanf("%lld",&t);while(t--){scanf("%s",s+1);n=strlen(s+1);ks=0;ans=0;marks=0;while(!left.empty()) left.pop();while(!mark.empty()) mark.pop();while(!k.empty()) k.pop();for(ll i=1;i<=n;i++){sum[i]=sum[i-1]+i;if(s[i]==')'&&left.empty())marks=ks=0;else if(s[i]==')'){ll z=(sum[i]-sum[left.top()-1])%XJQ;left.pop();marks=mark.top();mark.pop();ks=k.top();k.pop();ks=(ks+1)%XJQ;(marks+=z*ks)%=XJQ;(ans+=marks)%=XJQ;}else{left.push(i);mark.push(marks);k.push(ks);marks=0;ks=0;}}printf("%lld\n",ans);}
}

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T3 50分code

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=1e3+10;
int t,n,a[N],f[N][N*2];
int main()
{scanf("%d",&t);while(t--){scanf("%d",&n);a[n+1]=0;for(int i=n;i>=1;i--)scanf("%d",&a[i]);for(int i=1;i<=n;i++)a[i]+=a[i-1];memset(f,0x3f,sizeof(f));f[0][0]=0;for(int i=1;i<=n;i++){for(int j=n;j>=1;j--)f[i][j]=min(min(f[i][j*2-1],f[i][j*2])+a[i],f[i-1][j-1]);}printf("%d\n",f[n][1]);}
}

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总结

开心，题目难一点就翻车了$QVQ$。
$T1:$好像是$dfs$，就写了一个，然后发现有一个地方直接写了$return0$。
愉快爆蛋。
$T2:$写了半天，和正解檫肩而过，写了一个先加后$mod$就愉快30$，不如暴力$QVQ$
$T3:$明显$50$分$dp$
$QVQVQ$