NOIP2015 求和

 

 TM

 

#include<cstdio>
#define UP(a,b,c) for(int a=(b);a<=(c);a++)
using namespace std;
typedef long long LL;
const int N=100010,MOD=10007;
LL n,m,a[N],c[N],num[N][2],si[N][2],sn[N][2],ss[N][2],x;
int main(){
    scanf("%lld%lld",&n,&m);
    UP(i,1,n)scanf("%lld",&a[i]);
    UP(i,1,n)scanf("%lld",&c[i]);
    UP(i,1,n){
        num[c[i]][i&1]++;
        si[c[i]][i&1]+=i%MOD;
        sn[c[i]][i&1]+=a[i]%MOD;
        ss[c[i]][i&1]+=a[i]*i%MOD;
    }
    UP(i,1,m)UP(j,0,1)
        if(num[i][j]>1)x=(x+(si[i][j]*sn[i][j])%MOD+(num[i][j]-2)*ss[i][j])%MOD;
    printf("%lld\n",x);
    return 0;
}

　　

 

#include<cstdio>
#define FOR(a,b,c) for(int a=(b);a<=(c);a++)
using namespace std;
typedef long long LL;
const int maxn=100000+10;
const int MOD=10007;
LL n,m,a[maxn],c[maxn],sr[maxn][2],sa[maxn][2],s[maxn][2],sq[maxn][2];
int main(){
    scanf("%lld%lld",&n,&m);
    FOR(i,1,n)scanf("%lld",&a[i]);
    FOR(i,1,n)scanf("%lld",&c[i]);
    LL ans=0;
    FOR(i,1,n){
        int p=i&1,ci=c[i];
        s[ci][p]++;
         
        if(s[ci][p]>1)
             ans+=(a[i]*sr[ci][p])%MOD+(i*sa[ci][p])%MOD+sq[ci][p]+((s[ci][p]-1)*i*a[i]%MOD);
        ans%=MOD;
         
        sr[ci][p]+=i;
        sa[ci][p]+=a[i];
        sq[ci][p]+=i*a[i];
    }
    ans%=MOD;
    printf("%lld\n",ans);
    return 0;
}