1 条题解
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0
C :
#include <stdio.h> #include <stdlib.h> #include <string.h> int dp[110][110]; int main() { int n, m; while(scanf("%d%d",&n,&m) != EOF){ int A[110]; memset(dp,0,sizeof(dp)); for(int i = 1;i <= n; i++){ scanf("%d",&A[i]); } for(int i = 0;i <= A[1]; i++)dp[1][i] = 1; for(int i = 1;i <= n; i++)dp[i][0] = 1; for(int i = 2;i <= n; i++){ for(int j = 1;j <= m; j++){ for(int k = 0;k <= A[i]; k++){ if(j >= k)dp[i][j] = (dp[i][j] + dp[i-1][j-k])%1000007; } } } printf("%d\n",dp[n][m]); } return 0; }
C++ :
#include <iostream> using namespace std; const int N(109); int data[N]; int f[N][N]={0}; int main() { int n,m; cin>>n>>m; for(int i=1;i<=n;i++) cin>>data[i]; f[0][0]=1; for(int i=1;i<=n;i++) for(int j=0;j<=m;j++) { for(int k=0;k<=data[i];k++) { if(j-k<0) continue; f[i][j]+=f[i-1][j-k]; } f[i][j]=f[i][j]%1000007; } cout<<f[n][m]%1000007<<endl; return 0; }
Pascal :
var n,m,k,i,j:longint; f:array[0..101,0..101] of longint; a:array[1..101] of longint; begin fillchar(f,sizeof(f),0); readln(n,m); for i:=0 to n do f[i,0]:=1; for i:=1 to n do read(a[i]); for i:=1 to n do for j:=1 to m do for k:=0 to a[i] do if j-k>=0 then f[i,j]:=(f[i,j]+f[i-1,j-k]) mod 1000007; writeln(f[n,m]); end.
- 1
信息
- ID
- 762
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者