woj1010- Alternate Sum
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公式题,ans=(max*2^(n-1))%2006;
#include <iostream> #include<cstdio> using namespace std; const int MaxN = 1 << 30; int main() { int n; while(1 == scanf("%d", &n) && n ) { int maxx; scanf("%d", &maxx); for(int i = 0; i < n - 1; i++) { int t; scanf("%d", &t); maxx = max(maxx, t); } maxx %= 2006; if(maxx < 0) maxx += 2006; for(int i = 0; i < n - 1; i++) { maxx *= 2; maxx %= 2006; } printf("%d\n", maxx % 2006); } return 0; }
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