• 今日SGU 5.14


    //SGU 131 还没完全想清楚 留坑

    SGU 259

    题意:一个机器处理n个任务,每个任务有时间t和传送时间l

    收获:贪心

    #include<bits/stdc++.h>
    #define de(x) cout<<#x<<"="<<x<<endl;
    #define dd(x) cout<<#x<<"="<<x<<" ";
    #define rep(i,a,b) for(int i=a;i<(b);++i)
    #define repd(i,a,b) for(int i=a;i>=(b);--i)
    #define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
    #define ll long long
    #define mt(a,b) memset(a,b,sizeof(a))
    #define fi first
    #define se second
    #define inf 0x3f3f3f3f
    #define INF 0x3f3f3f3f3f3f3f3f
    #define pii pair<int,int>
    #define pdd pair<double,double>
    #define pdi pair<double,int>
    #define mp(u,v) make_pair(u,v)
    #define sz(a) (int)a.size()
    #define ull unsigned long long
    #define ll long long
    #define pb push_back
    #define PI acos(-1.0)
    #define qc std::ios::sync_with_stdio(false)
    #define db double
    #define all(a) a.begin(),a.end()
    const int mod = 1e9+7;
    const int maxn = 1e2+5;
    const double eps = 1e-6;
    using namespace std;
    bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
    bool ls(const db &a, const db &b) { return a + eps < b; }
    bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
    ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
    ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
    ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    ll read(){
        ll x=0,f=1;char ch=getchar();
        while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
        while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
        return x*f;
    }
    //inv[1]=1;
    //for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
    struct E{
        int t,l;
        bool operator<(const E& c)const{
            return l>c.l;
        }
    }e[maxn];
    int main(){
        int n;
        scanf("%d",&n);
        rep(i,0,n) scanf("%d",&e[i].t);
        rep(i,0,n) scanf("%d",&e[i].l);
        sort(e,e+n);
        int ans = -inf,tim = 0;
        rep(i,0,n){
            tim+=e[i].t;
            ans=max(ans,tim+e[i].l);
        }
        printf("%d
    ",ans);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/chinacwj/p/9038309.html
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