poj_1276

dp[i] 代表能否凑出总数为i的cash

#include <cstdio>
#include <cstring>

int n[11], d[11], cash, N;
bool dp[100010];

int main(int argc, char const *argv[])
{
    // freopen("in", "r", stdin);
    while(~scanf("%d%d", &cash, &N)){
        for(int i = 1; i <= N; ++i)
            scanf("%d %d", &n[i], &d[i]);
        memset(dp, false, sizeof(dp));
        dp[0] = true;
        for(int i = 1; i <= N; ++i)
            for(int j = cash; j >= 0; --j)
                if(dp[j] == true)
                    for(int k = 1; k <= n[i]; ++k){
                        if( j + k*d[i] > cash )
                            break;
                        dp[j + k*d[i]] = true;
                    }
        for(int i = cash; i >= 0; --i)
            if(dp[i] == true){
                printf("%d
", i);
                break;
            }
    }
    return 0;
}