題目描述

ftiasch 有 N 個物品, 體積分別是 W1, W2, …, WN。 由於她的疏忽, 第 i 個物品丟失了。 “要使用剩下的 N – 1 物品裝滿容積為 x 的背包，有幾種方法呢？” — 這是經典的問題了。她把答案記為 Count(i, x) ，想要得到所有1 <= i <= N, 1 <= x <= M的 Count(i, x) 表格。

輸入輸出格式

輸入格式：

第1行：兩個整數 N (1 ≤ N ≤ 2 × 10^3)N(1≤N≤2×103) 和 M (1 ≤ M ≤ 2 × 10^3)M(1≤M≤2×103)，物品的數量和最大的容積。

第2行： N 個整數 W1, W2, …, WN, 物品的體積。

輸出格式：

一個 N × M 的矩陣， Count(i, x)的末位數字。

This DP is pretty hard.

First we should know that F[i][j] means that how many funcation what we can have when we put i‘s stuff in the bag which has j‘s volume.

If the i‘s stuff had to taken, it wil be f[i-1][j-w[i]], else, it will be f[i-1][j], So we can get the funcation :f[i][j]=f[i-1][j]+f[i-1][j-w[i];

we can use rounded array change it to f[j]=f[j]+f[j-w[i]].

So, how can we get the count ?

we had to enumeration whitch stuff we had lost.

if w[i]>j, that means, all of the answer has include the stuff i, because it was bigger than the volume. Therefore, the answer should be f[j] , which means take all of the answer.

if w[i]<=j, that means there so anwer has be counted. what we should minus is the rest of stuff(except i) to pull in j-w[i]. which is f[j]-c[i][j-w[i]]

if w[i]==0 , the c[i][j] will be 1.

that‘s all.

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#define in(a) a=read()
#define REP(i,k,n)  for(int i=k;i<=n;i++) 
using namespace std;
inline int read(){
    int x=0,f=1;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar())
        if(ch==‘-‘)
            f=-1;
    for(;isdigit(ch);ch=getchar())
        x=x*10+ch-‘0‘;
    return x*f;
} 
int n,m;
int f[2010],c[2010][2010],w[2010]; 
int main(){
    in(n),in(m);
    REP(i,1,n)  in(w[i]); 
    f[0]=1;  
    REP(i,1,n)
        for(int j=m;j>=w[i];j--) 
            f[j]=(f[j]+f[j-w[i]])%10;
    REP(i,1,n){
        c[i][0]=1;
        REP(j,1,m){ 
            if(j<w[i])  c[i][j]=f[j];
            else  c[i][j]=(f[j]-c[i][j-w[i]]+10)%10;
            printf("%d",c[i][j]);
        }
        printf("\n");     
    } 
    return 0;
} 

洛谷P4141消失之物