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  • cf 383 D

    D. Antimatter
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Iahub accidentally discovered a secret lab. He found there n devices ordered in a line, numbered from 1 to n from left to right. Each device i (1 ≤ i ≤ n) can create either ai units of matter or ai units of antimatter.

    Iahub wants to choose some contiguous subarray of devices in the lab, specify the production mode for each of them (produce matter or antimatter) and finally take a photo of it. However he will be successful only if the amounts of matter and antimatter produced in the selected subarray will be the same (otherwise there would be overflowing matter or antimatter in the photo).

    You are requested to compute the number of different ways Iahub can successful take a photo. A photo is different than another if it represents another subarray, or if at least one device of the subarray is set to produce matter in one of the photos and antimatter in the other one.

    Input

    The first line contains an integer n (1 ≤ n ≤ 1000). The second line contains n integers a1a2, ..., an (1 ≤ ai ≤ 1000).

    The sum a1 + a2 + ... + an will be less than or equal to 10000.

    Output

    Output a single integer, the number of ways Iahub can take a photo, modulo 1000000007 (109 + 7).

    Sample test(s)
    input
    4
    1 1 1 1
    output
    12
    Note

    The possible photos are [1+, 2-], [1-, 2+], [2+, 3-], [2-, 3+], [3+, 4-], [3-, 4+], [1+, 2+, 3-, 4-], [1+, 2-, 3+, 4-], [1+, 2-, 3-, 4+], [1-, 2+, 3+, 4-], [1-, 2+, 3-, 4+] and [1-, 2-, 3+, 4+], where "i+" means that the i-th element produces matter, and "i-" means that the i-th element produces antimatter.

     dp 背包问题

     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 const int MAX_N = 1005;
     9 const int MOD = 1e9 + 7;
    10 int N,_max = 0;
    11 int a[MAX_N],dp[MAX_N][20005];
    12 
    13 void solve() {
    14         for(int i = 1; i <= N; ++i) {
    15                 for(int j = 2 * _max; j >= 0; --j) {
    16                         //dp[i][j] = dp[i - 1][j];
    17                         int v = 0;
    18                         if(j - a[i] == 0 + _max) v = 1;
    19                         dp[i][j] = (dp[i][j] + v + dp[i - 1][j - a[i]]) % MOD;
    20                         dp[i][j] = (dp[i][j] + v + dp[i - 1][j + a[i]]) % MOD;
    21 
    22                 }
    23         }
    24         int ans = 0;
    25        // printf("dds = %d %d
    ",dp[3][5],dp[3][3]);
    26         for(int i = 1; i <= N; ++i) ans = (ans + dp[i][0 + _max]) % MOD;
    27         printf("%d
    ",ans);
    28 
    29 }
    30 
    31 int main()
    32 {
    33     scanf("%d",&N);
    34     for(int i = 1; i <= N; ++i) {
    35             scanf("%d",&a[i]);
    36             //printf("%d 
    ",a[i]);
    37             _max += a[i];
    38     }
    39 
    40     solve();
    41     return 0;
    42 }
    View Code
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  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3669801.html
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