//#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inf 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
const int maxn=1e6+10;
using namespace std;
int dp[maxn];
int c[maxn];
int w[maxn];
int main()
{int n,m;while(cin>>n>>m){for(int i=0; i<n; ++i){cin>>c[i]>>w[i];}for(int i=0; i<n; ++i)for(int j=m; j>=c[i]; --j)dp[j]=max(dp[j],dp[j-c[i]]+w[i]);cout<<dp[m]<<endl;}return 0;
}
#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inf 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
const ll moad=1e9+7;
const int maxn=1e6+10;
int dp[maxn];
int w[maxn];
int vis[maxn];
int main()
{int t,n;cin>>t;while(t--){cin>>n;int sum=0;for(int i=0;i<n;++i)cin>>w[i],sum+=w[i];memset(dp,0,sizeof(dp));for(int i=0;i<n;++i){for(int j=sum/2;j>=w[i];--j){dp[j]=max(dp[j],dp[j-w[i]]+w[i]);}}int __=dp[sum/2],_;_=sum-__;cout<<abs(__-_)<<endl;}return 0;
}
#include <bits/stdc++.h>
#include <iostream>
#define X 10005
#define inf 0x3f3f3f3f
#define PI 3.141592653589793238462643383
#define IO ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
typedef long long ll;
const ll moad=1e9+7;
const int maxn=1e6+10;
int q[maxn];
int c[maxn];
int w[maxn];
int dp[maxn];
struct node
{double c,q,w;
}point[maxn];
bool cmp(node a,node b)
{//return a.q-a.c<b.q-b.c;//return a.c*b.q<b.c*a.q;return a.q/a.c<=b.q/b.c;
}
int main()
{int n,m;while(cin>>n>>m){memset(dp,0,sizeof(dp));for(int i=0;i<n;++i)cin>>point[i].c>>point[i].q>>point[i].w;sort(point,point+n,cmp);
// for(int i=0;i<n;++i)
// cout<<point[i].c<<' '<<point[i].q<<' '<<point[i].w<<endl;for(int i=0;i<n;++i){for(int j=m;j>=point[i].q;--j){dp[j]=max(dp[j],dp[j-int(point[i].c)]+int(point[i].w));}}cout<<dp[m]<<endl;}return 0;
}
poj1745
給你n個數，經過加減的操作，看能不能整除K。 dp[i][j]代表前i個數字整除 m 的余數為 j ?
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#define X 10005
using namespace std;
int main()
{int N,K;cin>>N>>K;int a[X];for(int i=1;i<=N;++i){scanf("%d",&a[i]);a[i]%=K;}int dp[X][K];memset(dp,0,sizeof(dp));//if(a[1]<0)// a[1]+=K;a[1]=(a[1]+K)%K;dp[1][a[1]]=1;for(int i=2;i<=N;++i){for(int j=0;j<K;++j){if(dp[i-1][j]){/* if(j+a[i]>=0)dp[i][j+a[i]]=1;elsedp[i][j+a[i]+K]=1;if(j>=a[i])dp[i][j-a[i]]=1;elsedp[i][j-a[i]+K]=1;*/dp[i][(j+a[i]+K)%K]=1;//保證下標大于0dp[i][(j-a[i]+K)%K]=1;}}}if(dp[N][0])cout<<"Divisible"<<endl;elsecout<<"Not divisible"<<endl;return 0;
}
?
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cstring>
#define X 10005
using namespace std;
long long POW(int a,int b)
{long long ans=1;while(b){if(b&1){ans*=a;}a*=a;b>>=1;}return ans;
}
int main()
{int N,K;cin>>N>>K;int a[X];for(int i=0;i<N;++i){scanf("%d",&a[i]);}long long n=POW(2,N-1);for(int i=0;i<n;++i){int sum=a[0];//cout<<a[0];for(int j=0;j<N-1;++j){if((i>>j)&1){//cout<<'-'<<a[j+1];sum+=a[j+1];}elsesum-=a[j+1];//,cout<<'+'<<a[j+1];}//cout<<'='<<sum<<"*****"<<sum%K<<endl;if(!(sum%K))//!的優先級比%的高{cout<<"Divisible"<<endl;return 0;}//cout<<endl;}cout<<"Not divisible"<<endl;return 0;
}
//這個方法是對的,可是在這兒N是10000算出來的n是2^10000會炸的
