Shenyang 2019icpc

【K题】Guanguan's Happy water

Meaning of the questions: in fact, seeking first k expectations of the current node

N for two cases:

1.n <= k ----- directly applied to a [1 ... n]

2.n>k----ans=suma+f[k+1...n]

Note that most of the data types which: important to note, all possible modulo modulus can

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=77;
const ll mod=1e9+7;
ll a[maxn],f[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    int k;
    ll n;
    while(t--){
        scanf("%d%lld",&k,&n);
        ll suma=0,sumf=0;
        for(int i=1;i<=k;i++){
            scanf("%lld",&a[i]);
            suma=(suma+a[i])%mod;
        }
        for(int i=1;i<=k;i++){
            scanf("%lld",&f[i]);
            sumf=(sumf+f[i])%mod;
        }
        ll ans=0;
        if(n<=k){
            for(inti = 1 ; i <= ( int ) n; i ++ ) 
            ans = (ANS + a [i])% v; 
        } 
        Else { 
            ans = suma% v;            
            ll cnt = (nk) / k;
            int len = cnt n-k * k; 
            cnt % = v; 
            ans = (cnt * sumf + ANS)% v;
             for ( int i = 1 ; i <= len; ++ i) 
                ans = (ANS + F [i])% v; 
        } 
        Printf ( " % lld \ n" , Year% change); 
    } 
}

 

//#pragma GCC optimize(2)

#include <bits/stdc++.h>
using namespace std;
#define _ 0
#define MX 100050
#define pb push_back
#define len length()
#define IO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);

typedef long long ll;
typedef pair <int,int> iip;
int i, j, cnt = 0;

ll dp[MX];

int main()
{
//    IO
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        dp[0] = 0;
        for(int i = 1; i <= m+10000; i++)
        {
            dp[i] = 0x1f1f1f1f1f1f1f1f;
        }
        for(int k = 0; k < n; k++)
        {
            ll p, c;
            scanf("%lld%lld", &p, &c);
            for(int i = c; i <= m+10000; i++)
            {
                dp[i] = min(dp[i-c]+p, dp[i]);
            }
        }
        ll a = 0x1f1f1f1f1f1f1f1f, b = m;
        for(int i = m; i <= m+10000; i++)
        {
            if(dp[i] <= a) a = dp[i], b = i;
        }
        printf("%lld %lld\n", a, b);
    }

    return ~~(0^_^0);
}