y总
#include <cstring>
#include <iostream>
using namespace std;
const int N = 1010, mod = 1e9 + 7;
int n, m;
int f[N], g[N];
int main()
{
cin >> n >> m;
memset(f, -0x3f, sizeof f);
f[0] = 0;
g[0] = 1;
for (int i = 0; i < n; i ++ )
{
int v, w;
cin >> v >> w;
for (int j = m; j >= v; j -- )
{
int maxv = max(f[j], f[j - v] + w);
int s = 0;
if (f[j] == maxv) s = g[j];
if (f[j - v] + w == maxv) s = (s + g[j - v]) % mod;
f[j] = maxv, g[j] = s;
}
}
int res = 0;
for (int i = 1; i <= m; i ++ )
if (f[i] > f[res])
res = i;
int sum = 0;
for (int i = 0; i <= m; i ++ )
if (f[i] == f[res])
sum = (sum + g[i]) % mod;
cout << sum << endl;
return 0;
}
稽宝
#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
const int N = 1010, mod = 1e9 + 7;
int f[N],cnt[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n,m;
cin >> n >> m;
for(int i = 0; i <= m; i ++ )
cnt[i] = 1;
for(int i = 0; i < n; i ++ )
{
int v,w;
cin >> v >> w;
for(int j = m; j >= v; j -- )
{
int maxx = f[j - v] + w;
if(maxx > f[j]) //价值最大,选其求方案数
{
f[j] = maxx;
cnt[j] = cnt[j - v] % mod;
}
else if (maxx == f[j]) //相等, 更新两个的总和为方案数
{
cnt[j] = (cnt[j] + cnt[j - v]) % mod;
}
}
}
cout << cnt[m] << "\n";
return 0;
}