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摘要： #include <iostream>#include <cstdio>#include <string.h>using namespace std;int dp[15000],t,sum;/*void bagg01(int value,int cost){ for (int i=t;i>=cost;--i) dp[i]=(dp[i]+dp[i-cost])%10000;}void baggall(int value,int cost){ for (int i=cost;i<=t;++i) dp[i]=(dp[i]+dp[i-cost])%100 阅读全文