2018_5_12 模拟赛

前言

水了……

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SSL 2413 2414 2415

NO.1 NO.2 NO.3 排名 简写单词 连通块

排名（水而玄学，直接贴代码）

#include <cstdio>
#include <algorithm>
using namespace std;
struct score{int chi,mat,eng,itx,rank;}sco[51]; int n;
bool cmp(score x,score y){return x.mat>y.mat;}
bool cmp2(score x,score y){return x.itx>y.itx;}
int main(){
    freopen("sort.in","r",stdin);
    freopen("sort.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    scanf("%5d%5d%5d%5d",&sco[i].chi,&sco[i].mat,&sco[i].eng,&sco[i].itx);
    stable_sort(sco+1,sco+1+n,cmp);
    for (int i=1;i<=n;i++) sco[i].rank=i;
    stable_sort(sco+1,sco+1+n,cmp2);
    for (int i=1;i<=n;i++)
    printf("%4d%5d%5d%5d%5d%5d \n",sco[i].chi,sco[i].mat,sco[i].eng,sco[i].itx,sco[i].rank,i);//玄学输出
    return 0;
}

简写单词（我可能是……了，代码）

#include <cstdio>
#include <cstring>
#define p 3001
using namespace std;
int n; char s[51][51]; bool v[51];
int main(){
    freopen("abbreviate.in","r",stdin);
    freopen("abbreviate.out","w",stdout);
    scanf("%d\n",&n);
    for (int i=1;i<=n;i++) scanf("%s",s[i]);
    for (int i=1;i<=n;i++){
    char st[51]=""; memset(v,0,sizeof(v));
    for (int j=0;j<strlen(s[i]);j++){
        int sum=0; st[j]=s[i][j];
        for (int k=1;k<=n;k++)
        if (st[j]!=s[k][j]||v[k]) v[k]=1,sum++;//hate判重
        if (sum!=n-1) continue;
        puts(st); break;
    }
    }
    return 0;
}

连通块（水水的并查集（MLE了））

#include <cstdio>
using namespace std;
bool a[501][501],b[501][501]; int ans,n,m,f[250001],sum;
int getf(int u){return (f[u]==u)?u:f[u]=getf(f[u]);}
void uni(int x1,int y1,int x2,int y2){
    if (!a[x2][y2]||b[x1][y1]!=b[x2][y2]) return;//未出现棋子或者棋子颜色不一样
    int fa=getf((x1-1)*n+y1),fb=getf((x2-1)*n+y2);
    if (fa!=fb) sum++,f[fa]=fb;
}
int main(){
    freopen("blocks.in","r",stdin);
    freopen("blocks.out","w",stdout);
    scanf("%d%d",&n,&m);
    for (int i=1;i<=n;i++)
    for (int j=1;j<=n;j++)
    f[(i-1)*n+j]=(i-1)*n+j;
    for (int i=1;i<=m;i++){
        int q,x,y; sum=0;
        scanf("%d%d%d",&q,&x,&y);
        a[x][y]=1; b[x][y]=q;
        uni(x,y,x-1,y); uni(x,y,x+1,y);
        uni(x,y,x,y+1); uni(x,y,x,y-1);//四个方向
        ans=ans-sum+1;//连到多少
        printf("%d\n",ans);
    }
    return 0;
}

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SSL NO.4 2416 条形图

题目

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分析

dp，状态转移方程：f[i][j]表示第i行有j个棋子的方案。
$f[i][j]=f[i-1][j]+f[i][j-1]$
最后算出$f[n][1]+……+f[n][n]-1就好了（避免空）$
高精度，这是最重要的。

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代码

#include <cstdio>
using namespace std;
char f[101][101][61]; short n;
void add(char *a,char *b,char *c){
    int g=0;
    for (int i=1;i<=60;i++){
        a[i]=g+b[i]+c[i];
        g=a[i]/10;
        a[i]%=10;
    }
}
int main(){
    freopen("diagrams.in","r",stdin);
    freopen("diagrams.out","w",stdout);
    scanf("%d",&n);
    for (int i=1;i<=n;i++) f[i][0][1]=1;
    for (int i=1;i<=n;i++)
    for (int j=1;j<=i;j++) 
    add(f[i][j],f[i-1][j],f[i][j-1]);
    f[n][1][1]--;
    for (int i=1;i<=n;i++) 
    add(f[n][i],f[n][i],f[n][i-1]);
    int k=60;
    while (!f[n][n][k]) k--;
    for (int i=k;i>=1;i--) putchar(f[n][n][i]+48);
    return 0;
}