noip2012 普及组

T1 质因数分解 题目传送门

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define LL long long 
using namespace std;
LL read(){
    LL ans=0,f=1,c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
    return ans*f;
}
LL n;
LL find(LL n){
    int x=2;
    while(x<=n&&n%x) x++;
    return x;
}
int main()
{
    n=read();
    printf("%lld
",n/find(n));
    return 0;
}
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T2 寻宝 题目传送门

#include<cstdio>
#include<cstring>
int a[10005][105],k[10005][105],c[10005];
int main()
{
    int i,j,n,m,sum=0,go,ha,s;
    scanf("%d %d",&n,&m);
    for(i=1;i<=n;i++)
    for(j=0;j<=m-1;j++)
    {
        scanf("%d %d",&a[i][j],&k[i][j]);
        if(a[i][j]==1)
        c[i]++;
    }
    scanf("%d",&go);
    for(i=1;i<=n;i++)
    {    
        sum=(sum+k[i][go])%20123;
        s=(k[i][go]-1)%c[i]+1;
        if(a[i][go]==1) ha=1;
        else ha=0;
        while(ha<s)
        {
            if(go==m-1) go=0;
            else go++;
            if(a[i][go]==1)
            ha++;
        }
    }
    printf("%d
",sum);
    return 0;
}
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T3 摆花 题目传送门

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int M=1007,mod=1000007;
int read(){
    int ans=0,f=1,c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
    return ans*f;
}
int n,m,f[M][M],w[M];
int main()
{
    n=read(); m=read();
    for(int i=1;i<=n;i++) w[i]=read();
    for(int i=0;i<=n;i++) f[i][0]=1;
    for(int i=1;i<=n;i++)
     for(int j=1;j<=m;j++){
         f[i][j]=f[i-1][j]+f[i][j-1];
         if(j-w[i]) f[i][j]-=f[i-1][j-w[i]-1];
         f[i][j]=(f[i][j]+mod)%mod;
     }
    printf("%d
",f[n][m]);
    return 0;
}
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T4 文化之旅 题目传送门

这道题还是有点需要说的 我写的是A* 先一波spfa预处理如果没有限制需要的路程

然后在有限制的跑一波dfs就好了

#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int M=155,inf=0x3f3f3f3f;
int read(){
    LL ans=0,f=1,c=getchar();
    while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();}
    while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();}
    return ans*f;
}
int n,k,m,S,T,ans=inf,cnt;
int w[M],d[M][M],map[M][M],dis[M],q[M],vis[M],usd[M];
void spfa(){
    memset(dis,0x3f,sizeof(dis));
    int head=0,tail=1;
    q[0]=T; vis[T]=1; dis[T]=0;
    while(head!=tail){
        int x=q[head++]; vis[x]=0;if(head>M) head=0;
        for(int now=1;now<=n;now++){
            if(dis[now]>dis[x]+d[now][x]){
                dis[now]=dis[x]+d[now][x];
                if(!vis[now]){q[tail++]=now; vis[now]=1; if(tail>M) tail=0;}
            }
        }
    }
}
int check(int x){
    for(int i=1;i<=cnt;i++) if(map[w[x]][usd[i]]||usd[i]==w[x]) return 0;
    return 1;
}
void dfs(int x,int h){
    if(x==T){ans=min(ans,h); return ;}
    for(int now=1;now<=n;now++){
        if(vis[now]||h+dis[now]>=ans||!check(now)) continue;
        vis[now]=1;
        usd[++cnt]=w[now];
        dfs(now,h+d[x][now]);
        cnt--; vis[now]=0;
    }
}
int main()
{
    int x,y,v;
    memset(d,0x3f,sizeof(d));
    n=read(); k=read(); m=read();
    S=read(); T=read();
    for(int i=1;i<=n;i++) w[i]=read();
    for(int i=1;i<=k;i++)
     for(int j=1;j<=k;j++)
      map[i][j]=read();
    for(int i=1;i<=m;i++){
        x=read(); y=read(); v=read();
        if(!map[w[x]][w[y]]) d[y][x]=min(d[y][x],v);
        if(!map[w[y]][w[x]]) d[x][y]=min(d[x][y],v);
    }
    spfa();
    vis[S]=1; usd[++cnt]=w[S]; dfs(S,0);
    if(ans<inf) printf("%d
",ans);
    else printf("-1
");
    return 0;
} 
View Code
原文地址:https://www.cnblogs.com/lyzuikeai/p/7247547.html